Engineering Pure Maths IB

Rao W. O.

Contents

Course description ivReferences iv

Chapter 1. Derivatives of trigonometric functions 1

Chapter 2. Applications of Diferentiation 62.1. Approximating Small Changes 62.2. Mean value theorems of differential calculus 82.3. Indeterminate Forms 112.4. Extreme Values 112.5. Concavity and Inflections 182.6. Curve Sketching 212.7. Linear Approximations 262.8. Taylor Polynomials 26

Chapter 3. Integration 273.1. Antiderivatives 273.2. The Definite Integral 283.3. The Fundamental Theorem of Calculus 31

Chapter 4. Methods of Integration 344.1. Method of Substitution 344.2. Integration by Parts 364.3. Reduction Formulas 384.4. Integrals of Rational Functions 404.5. Inverse Substitution 454.6. Improper Integrals 49

Chapter 5. Applications of Integration 555.1. Areas of Plane Figures 555.2. Volume 575.3. Arc Length 605.4. Areas of Surfaces of Revolution 63

Chapter 6. Polar Coordinates 666.1. Polar Coordinates and Polar Curves 666.2. Slopes for Polar Curves 686.3. Area Bounded by Polar Curves 70

iii

Course description

Applications of Differential CalculusTrigonometric functions; their derivatives, stationary points, minima and maxima prob-lems. Increasing and decreasing functions. Small increments, tangents. Applications ofdifferentiation to curve sketching; convex inflexion points. Polar co-ordinates.Integral CalculusIntegration as inverse of differentiation. Definite integral as limit of sum. Standard forms.Applications of integration to area, volume.

References

1. Calculus: A complete course by Robert A. Adams and Christopher Essex.2. Engineering Mathematics by K. A. Stroud.

The proposed course outline is as follows

1. Week one• Derivatives of trigonometric functions.• Approximating small changes.

2. Week two• Extreme values.

3. Week three• Mean value theorem of differential calculus.• Concavity and inflections.• Curve sketching.

4. Week four• Antiderivatives.• The definite integrals.

5. Week five• The fundamental theorem of calculus.• Method of substitution.• Integration by parts

6. Week six• Reduction formulas.• Integrals of rational functions.

7. Week seven• Inverse substitution.

8. Week eight• Areas of plane figures.• Volume.

9. Week nine• Arc length.• Areas of surfaces of revolution.

10. Week ten• Polar coordinates and polar curves.• Slopes for polar curves.• Areas bounded by polar curves

iv

CHAPTER 1

Derivatives of trigonometric functions

Trigonometric functions, especially sine and cosine arise whenever quantities fluctuatein a periodic way. We begin by establishing the following limits

Theorem 1.1. The functions of sin θ and cos θ are continuous at every value of θ. Inparticular, we have

limθ→0

sin θ = 0 and limθ→0

cos θ = 1.

The result of the above theorem is obvious from the graphs of sine and cosine.

Theorem 1.2.

limθ→0

sin θ

θ= 1 (θ is in radians) .

Proof. Let 0 < θ < π2 and represent θ as in the diagram above. Now

Area 4OAP < Area sector OAP < Area 4OAT.

Area 4OAP =1

2· (1) sin θ =

sin θ

2.

Area sector OAP =θ

2π· π · (1)2 =

θ

2.

O

P = (cos θ, sin θ)

T = (1, tan θ)

A = (1, 0)

y

x

1

θ

Figure 1.1. Area 4OAP <Area sector OAP < Area 4OAT

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PURE MATHS IB

Area 4OAT =1

2· (1) tan θ =

sin θ

2 cos θ.

Thus

sin θ

2<θ

2<

sin θ

2 cos θ.

Multiplying through by the positive number 2sin θ we obtain

1 <θ

sin θ<

1

cos θ.

Taking the reciprocals we have

1 >sin θ

θ> cos θ.

Since

limθ→0

1 = 1 and limθ→0

cos θ = 1

it follows by squeeze (sandwich) theorem that

limθ→0

sin θ

θ= 1.

�

Example 1.1.1.Show that

limh→0

cosh− 1

h= 0.

Solution. Using half-angle formula

cosh = 1− 2 sin2

(h

2

),

we have that

limh→0

cosh− 1

h= lim

h→0−2 sin2 (h/2)

h

=

[θ = h/2

⇒ h = 2θ

]= − lim

θ→0

(2

sin θ

2θ· sin θ

)= −1 · 0 = 0

Theorem 1.3.

d

dxsinx = cosx.

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Proof.

d

dxsinx = lim

h→0

sin (x+ h)− sinx

h

= limh→0

sinx cosh+ cosx sinh− sinx

h

= limh→0

sinx (cosh− 1) + cosx sinh

h

= limh→0

sinx (cosh− 1)

h+ limh→0

cosx sinh

h

= sinx (0) + cosx (1) = cosx.

�

Theorem 1.4.d

dxcosx = − sinx.

Proof. Using the complementary angle identities, sin ((π/2)− x) = cosx andcos ((π/2)− x) = sinx and the chain rule we have

d

dxcosx =

d

dxsin(π

2− x)

= (−1) cos(π

2− x)

= −1{

cosπ

2cosx+ sin

π

2sinx

}= − sinx.

�

Example 1.1.2.Evaluate the derivatives of the following functions

(a) sin (πx) + cos (3x) (b) x2 sin√x (c) cosx

1−sinx

Solution.

(a) By sum and chain rules:

d

dx(sin (πx) + cos (3x)) = π cos (πx)− 3 sin (3x)

(b) By product and chain rules:

d

dx

(x2 sin

√x)

= 2x sin√x+ x2

1

2x−

12 cos

√x = 2x sin

√x+

1

2x

32 cos

√x.

(c) By quotient rule

d

dx

cosx

1− sinx=

(− sinx) (1− sinx)− cosx (− cosx)

(1− sinx)2

=− sinx+ sin2 x+ cos2 x

(1− sinx)2=

1

1− sinx

The following derivatives can be easily verified

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d

dxtanx = sec2 x

d

dxsecx = secx tanx

d

dxcotx = − csc2 x

d

dxcscx = − cscx cotx.

Example 1.1.3.Find the tangent and the normal lines to the curve y = tan

(πx4

)at the point (1, 1).

Solution. We have that

dy

dx=(

sec2(πx

4

))(π4

).

Therefore

dy

dx

∣∣∣∣x=1

=π

4sec2

(π4

)= · · · = π

2.

The tangent line is

y − 1

x− 1=π

2which simplifies to y = 1 +

π

2(x− 1) .

The normal line is

y − 1

x− 1= − 2

πwhich simplifies to y = 1− 2

π(x− 1) .

Exercise 1.1.

1. Verify the formula for the derivative of

(a) cscx. (b) cotx.

2. Find the derivative of the functions below. Simplify your answer whenever pos-sible.

(a) y = cos (3x).

(b) y = tan (πx)

(c) y = sin3(πx2

).

(d) y = sinx

5.

3. Given that sin 2x = 2 sinx cosx, deduce that cos 2x = cos2 x− sin2 x.4. Given that cos 2x = cos2 x− sin2, deduce that sin 2x = 2 sinx cosx.5. Find equations of the tangent and normal lines to the curve y =

√2 cos (x/4) at

(π, 1).6. Find an equation of the tangent line to the curve y = sinx◦ at where x = 45.7. Evaluate the following limits

(a) limx→0

(x2 cscx cotx

).

(b) limx→0

tan (2x)

x.

(c) limh→0

1− cosh

h2.

Solution.

1.

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(a) − cscx cotx. (b) − cscx.

2. (a) Ans −3 sin 3x.

(b) π sec2 πx.

(c)3

2sin2

(πx2

)cos(πx

2

).

(d)1

5cos

x

5.

5. Tangent line 4y = x− π + 4, Normal line y = 4π + 1− 4x.

6. y =1√2

+π

180√

2(x− 45)

7. (a) 1. (b) 2.

5

CHAPTER 2

Applications of Diferentiation

2.1. Approximating Small Changes

Suppose dx is regarded as a new independent variable called the differential of x,we can define a new dependent variable dy, called the differential of y as a function ofx and dx by

dy =dy

dxdx = f ′ (x) dx.

For example if y = x2, then dy = 2xdx, meaning the same thing as dy/dx = 2x. Iff (x) = 1/x, then df (x) = −

(1/x2

)dx.

If y is a function of x, y = f (x), then denoting a small change in x by dx instead of∆x, the corresponding small change in y, ∆y is approximated by the differential dy, i.e.

∆y ≈ dy = f ′ (x) dx.

Example 2.1.1.Without using a scientific calculator, determine by approximately how much the value ofsinx increases as x increases from π/3 to (π/3) + 0.006. To 3 decimal places, what is thevalue of sin ((π/3) + 0.006)?

graph of f

dy

∆y

y = f (x)

y

dx = ∆x

xx x+ dx

Figure 2.1. dy, the change in height to tangent line, approximates ∆y, the

change in height to the graph f

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PURE MATHS IB

Solution. If y = sinx, then dydx = cosx. Now x = π

3 ≈ 1.0472 and dx = 0.006.Therefore

dy =dy

dxdx = cosxdx = cos

(π3

)· 0.006 =

1

2(0.006) = 0.003.

This means that the change in the value of sinx is approximately 0.003. Consequently

sin((π

3

)+ 0.006

)≈ sin

(π3

)+ 0.003 = · · · = 0.869(3 d. p.).

Suppose changes in x are measured with respect to the size of x, then

relative change in x =dx

xand percentage change in x = 100

dx

x.

Example 2.1.2.By approximately what percentage does the area of a circle increases if the radius increasesby 2%?

Solution. A = πr2 implying that dAdr = 2πr. Now ∆A ≈ dA = 2πrdr and relative

change in A is

∆A

A≈ dA

A=

2πrdr

πr2= 2

dr

r.

If r increases by 2% then dr = 2100r, so

dA

A= 2 · 1

r· 2

100· r =

4

100

so that the area increases by 4%.

Exercise 2.1.

1. Use differentials to determine approximate change in the values of the givenfunction as its argument changes from the given value to the given amount.What is the approximate value of the function after the change?(a) y = 1/x as x increases from 2 to 2.01.(b) h (t) = cos (πt/4) as t increases from 2 to 2 + (1/10π).(c) f (x) =

√3x+ 1, as x increases from 1 to 1.08.

2. Find the approximate percentage changes in the given function that will resultfrom an increase of 2% in the value of x.

(a) y = x2 (b) y = 1/x2 (c) y = x3.

3. By approximately what percentage will the volume of a ball of radius r increaseif the radius increases by 2%?

4. What is the approximate value of f (x) =√

3x+ 1 after x has increased from 1to 1.08.

Solution.

1. (a) −0.0025, 0.4975.

(b) − 1

40, − 1

40.

(c) 0.06, 2.06.2.

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PURE MATHS IB

(a) 4%. (b) −4%. (c) 6%.

3. 6%.4. 2.02.

2.2. Mean value theorems of differential calculus

Theorem 2.1 (Mean Value Theorem). Suppose that the function f is continuous onthe closed finite interval [a, b] and that it is differentiable on the interval (a, b). Then ∃ apoint c ∈ (a, b) such that

f (b)− f (a)

b− a= f ′ (c) .

It means that the slope of the chord joining the points (a, f (a)) and (b, f (b)) is equal tothe slope of the tangent line to te curve y = f (x) at the point (c, f (c)) so that the twolines are parallel.

a

A (a, f (a))

C

y = f (x)

y

c x

B (b, f (b))

b

Figure 2.2. There is a point C on the curve where the tangent is parallel to

the chord AB

Example 2.2.1.Verify the conclusion of the mean value theorem for f (x) =

√x on the interval [a, b],

where a ≤ x ≤ b.

Solution. We are to show that ∃ c ∈ (a, b) such that

f (b)− f (a)

b− a= f ′ (c)

so long as f is continuous on [a, b] and is differentiable on (a, b). Now f ′ (x) = 12√x

,

f (a) =√a, f (b) =

√b.

∴1

2√c

=

√b−√a

b− a=

√b−√a(√

b−√a)(√

b+√a) =

1√b+√a.

The above equality implies that√c =

√b+√a

2 so that c =(√

b+√a

2

)2. Since a < b, we have

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PURE MATHS IB

a =

(√a+√a

2

)2

<

(√b+√a

2

)2

<

(√b+√b

2

)2

= b

which implies that c ∈ (a, b).

Example 2.2.2.Show that sinx < x for all x > 0.

Solution. If x > 2π, then sinx ≤ 1 < 2π < x. If 0 < x ≤ 2π, then by MVT, ∃c ∈ (0, x) such that

sinx

x=

sinx− sin 0

x− 0= [MVT on [0, x]] =

d

dxsinx

∣∣∣∣x=c

= cos c < 1

which implies that sinx < x in this case too.

Definition 2.2.1 (Increasing and decreasing functions). increasingdecreasingfunctions

Suppose that the function fis defined on an interval I and that x1 and x2 are two points in I.

(a) If f (x2) > f (x1) whenever x2 > x1, we say that f is increasing on I.(b) If f (x2) < f (x1) whenever x2 > x1, we say that f is decreasing on I.(c) If f (x2) ≥ f (x1) whenever x2 > x1, we say that f is non-decreasing on I.(d) If f (x2) ≤ f (x1) whenever x2 > x1, we say that f is non-increasing on I.

Diagram Fig 2.31

Theorem 2.2. derivative ofincreasing anddecreasingfunctions

Let J be an open interval and let I be an interval consisting of allpoints in J and possibly one or both of the end points of J . Suppose that f is continuouson I and differentiable on J .

(a) If f ′ (x) > 0 for all x ∈ J , then f is increasing on I.(b) If f ′ (x) < 0 for all x ∈ J , then f is decreasing on I.(c) If f ′ (x) ≥ 0 for all x ∈ J , then f is non-decreasing on I.(d) If f ′ (x) ≤ 0 for all x ∈ J , then f is non-increasing on I.

Example 2.2.3.On what intervals is the function f (x) = x3 − 12x+ 1 increasing? On what intervals is itdecreasing?

Solution. f ′ (x) = 3x2 − 12 = 3 (x− 2) (x+ 2). It follows that f ′ (x) > 0 whenx < −2 or x > 2 and f ′ (x) < 0 when −2 < x < 2. Therefore f is increasing on theintervals (−∞,−2) and (2,∞) and is decreasing on the interval (−2, 2).

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PURE MATHS IB

y = x3 − 12x+ 1

x

y

(−2, 17)

(2,−15)

Figure 2.3. The graph y = x3 − 12x+ 1

Example 2.2.4.Show that f (x) = x3 is increasing on any interval.

Solution. Let x1, x2 be any two real numbers satisfying x1 < x2. Since f ′ (x) =3x2 > 0 for all x 6= 0, we have that f (x1) < f (x2) if either x1 < x2 ≤ 0 or 0 ≤ x1 < x2.If x1 < 0 < x2, then f (x1) < 0 < f (x2). Thus f is increasing on every interval.

Exercise 2.2.

1. Illustrate the MVT by finding any points in the open interval (a, b) where thetangent line is parallel to the chord joining (a, f (a)) and (b, f (b)).(a) f (x) = x2 on [a, b].(b) f (x) = x3 − 3x+ 1 on [−2, 2].

(c) f (x) =1

xon [1, 2].

2. Show that tanx > x for 0 < x < π2 .

3. Let r > 1. If x > 0 or −1 ≤ x < 0, show that (1 + x)r < 1 + rx.4. Let 0 < r < 1. If x > 0 or −1 ≤ x < 0, show that (1 + x)r < 1 + rx.5. Find intervals of increase and decrease of the following functions

(a) f (x) = x3 − 4x+ 1.

(b) f (x) =(x2 − 4

)2.

(c) f (x) = x3 (5− x)2.(d) f (x) = x+ sinx.(e) f (x) = x2 + 2x+ 2.(f) f (x) = x3 − 4x+ 1.

6. Let f (x) =

x+ 2x2 sin (1/x) if x 6= 0,

0 if x = 0.

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PURE MATHS IB

(a) Show that f ; (0) = 1. (Hint: Use the definition of the derivative.)

(b) Show that any interval containing x = 0 also contains points where f ′ (x) <0, so f cannot be increasing on such an interval.

7. Show that the graphs of y = secx, and y = cscx have horizontal tangents andidentify those points.

Solution.

1. (a) c = b+a2 .

(b) c = ± 2√3.

(c) c =√

25. Find intervals of increase and decrease of the following functions

(a) Increasing on(−∞,− 2√

3

)and

(2√3,∞)

; decreasing on(− 2√

3, 2√

3

).

(b) Increasing on (−2, 0) and (2,∞); decreasing on (−∞,−2) and (0, 2).(c) Increasing on (−∞, 3) and (5,∞); decreasing on (3, 5).(d) Increasing on (−∞,∞).(e) Decreasing on (−∞,−1) and increasing on (−1,∞)

(f) Increasing on

(−∞,− 2√

3

)and on

(2√3,∞)

while it is decreasing on(− 2√

3,

2√3

)2.3. Indeterminate Forms

2.4. Extreme Values

2.4.1. Maximum and Minimum Values.

Definition 2.4.1. Function f has an absolute maximum value f (x0) at x0 in itsdomain if f (x) ≤ f (x0) holds for every x in the domain of f .

Similarly, f has an absolute minimum value f (x0) at x0 in its domain if f (x) ≥f (x0) holds for every x in the domain of f .

Remark 1.

1. A function will have only one absolute maximum (or minimum) value if it exists.However the value can occur at many points. For example, f (x) = sinx hasabsolute maximum of 1 but it occurs at every point π

2 + 2nπ, n ∈ Z.

2. A function need not have any extreme value. The function f (x) = 1x becomes

arbitrarily large as x approaches 0 from the right, and so has no finite absolutemaximum value.

Theorem 2.3. If the domain of the function f is a closed, finite interval or a unionof finitely many such intervals, and if f is continuous on that domain, then f must havean absolute maximum value and absolute minimum value.

Definition 2.4.2. Function f has a local maximum value (loc. max.) f (x0) atthe point x0 in its domain provided ∃ a number h > 0 such that f (x) ≤ f (x0) wheneverx is in the domain of f and |x− x0| < h.

Similarly, f has a local minimum value (loc. min.) f (x1) at the point x1 in itsdomain provided ∃ a number h > 0 such that f (x) ≥ f (x1) whenever x is in the domainof f and |x− x1| < h.

From the above figure we see that local extreme values can occur at any of the followingpoints.

(i) critical points of f ; points x ∈ D (f) where f ′ (x) = 0.

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y = f (x)

a x1 x2 x3 x4 x5 x6 b x

y

Figure 2.4. Local extreme values

(ii) singular points of f ; points x ∈ D (f) where f ′ (x) is not defined.(iii) endpoints of the domain of f ; points tha do belong to D (f) but are not interior

points of D (f).

In the figure above, x1, x3, x4 and x6 are critical points, x2 and x5 are singular pointsand a and b are endpoints.

Theorem 2.4. If the function f is defined on an interval I and has a local maximum(or local minimum) value at the point x = x0 in I, then x0 must be either a critical pointof f , a singular point of f , or an endpoint of I.

Example 2.4.1.Find the maximum and minimum values of the function g (x) = x3 − 3x2 − 9x+ 2 on theinterval −2 ≤ x ≤ 2.

Solution. Since g is a polynomial it can never have a singular point. For criticalpoints we calculate

g′ (x) = 3x2 − 6x− 9 = 3 (x+ 1) (x− 3) = 0.

Thus the critical points are at x = −1 or x = 3. But x = 3 is not in the domain of g andso we ignore it. We then investigate the endpoints x = −2 and x = 2 and critical pointx = −1.

g (−2) = 0, g (−1) = 7, g (2) = −20The maximum value of g on −2 ≤ x ≤ 2 is at the critical point x = −1, and the

minimum value is at the endpoint x = 2.

Example 2.4.2.Find and classify the local and absolute extreme values of the function f (x) = x − x2/3on the interval [−1, 2]. Sketch the graph.

Solution. f ′ (x) = 1 − 2

3x−1/3 =

(x1/3 − 2

3

)/x1/3. There is a singular point at

x = 0, and a critical point at x = 8/27. The end points are at x = −1 and x = 2.

f (−1) = −2, f (0) = 0, f (8/27) = −4/27, f (2) = 2− 22/3 ≈ 0.4126.

Tabular summary of the data on the graph is as followsThere are two local minima and two local maxima. The absolute maximum of f is f (2) =

2− 22/3 and the absolute minimum is f (−1) = −2.

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(−2, 0)

(−1, 7)

(2,−20)

y = g (x)

= x3 − 3x2 − 9x+ 2

x

y

Figure 2.5. g has maximum and minimum values 7 and −20, respectively

EP SP CP EPx −1 0 8/27 2

f ′ + undef − 0 +f min ↗ max ↘ min ↗ max

(−1,−2)

(827 ,

−427

)(2, 2− 22/3

)

y = x− x2/3x

y

Figure 2.6. The graph y = x− x2/3

Theorem 2.5 (The first derivative test). Part I: Testing interior critical points andsingular points.Suppose f is continuous at x0, and x0 is not an endpoint of the domain of f .(a) If ∃ an open interval (a, b) containing x0 such that f ′ (x) > 0 on (a, x0) and

f ′ (x) < 0 on (x0, b), then f has a local maximum value at x0.(b) If ∃ an open interval (a, b) containing x0 such that f ′ (x) < 0 on (a, x0) and

f ′ (x) > 0 on (x0, b), then f has a local minimum value at x0.Part II: Testing endpoints of the domain.

Suppose a is a left endpoint of the domain of f and f is right continuous at a.(c) If f ′ (x) > 0 on some interval (a, b) , then f has a local minimum value at

a.(d) If f ′ (x) < 0 on some interval (a, b) , then f has a local maximum value at

a.Suppose b is a right endpoint of the domain of f and f is left continuous at b.

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(e) If f ′ (x) > 0 on some interval (a, b) , then f has a local maximum value atb.

(f) If f ′ (x) < 0 on some interval (a, b) , then f has a local minimum value atb.

Remark 2.If f ′ is positive (or negative) on both sides of a critical or singular point, then f has neithera maximum nor a minimum value at that point.

Example 2.4.3.Find the local and absolute extreme values of f (x) = x4 − 2x2 − 3 on the interval [−2, 2].Sketch the graph of f .

Solution. f ′ (x) = 4x3 − 4x = 4x(x2 − 1

)= 4x (x+ 1) (x− 1).

The critical points are x = −1, x = 0, x = 1. The corresponding values are f (−1) =−4, f (0) = −3, f (1) = −4. There are no singular points. The values of f at the endpoints−2 and 2 are f (−2) = −5 and f (2) = 5.We summarize the positive/negative propertiesof f (x) and the implied increasing/decreasing behaviour of f (x) in a chart form below;

EP CP CP CP EPx −2 −1 0 1 2

f ′ − 0 + 0 − 0 +f max ↘ min ↗ max ↘ min ↗ max

(−2, 5)

(−1,−4)

−3

(2, 5)

(1,−4)

x

y

Figure 2.7. The graph y = x4 − 2x2 − 3

Theorem 2.3 can be adapted for open intervals as follows.

Theorem 2.6 (Existence of extreme values on open intervals). If f is continuous onan open interval (a, b), and if

limx→a+

f (x) = L and limx→b−

f (x) = M

then the following conclusions hold

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(i) If f (u) > L and f (u) > M for some u ∈ (a, b), then f has an absolute maximumvalue on (a, b).

(ii) If f (u) < L and f (u) < M for some u ∈ (a, b), then f has an absolute minimumvalue on (a, b).

In this theorem a may be −∞ and b may ∞. Also, either or both L and M may be∞ or −∞.

Example 2.4.4.Show that f (x) = x+ 4

x has an absolute minimum value on the interval (0,∞), and findthat minimum value.

Solution. We have

limx→0+

f (x) =∞ and limx→∞

f (x) =∞.

Since f (1) = 5 < ∞, Theorem 2.6 guarantees that f must have an absolute minimumvalue at some point in (0,∞). Now

f ′ (x) = 1− 4

x2=x2 − 4

x2=

(x− 2) (x+ 2)

x2

which equals 0 when x = ±2. Since D (f) = (0,∞) it has no singular points but has onlyone critical point at x = 2. Here f (2) = 4. This must be the minimum value of f on(0,∞).

(2, 4)

y = x+ 4x

x

y

Figure 2.8. f has minimum value 4 at x = 2

2.4.2. Extreme-Value Problems.Checklist for extreme-value problems

1. Make a diagram if appropriate.2. Define any symbols you wish to use that are not already specified in the statement

of the problem.3. Express the quantity Q to be maximized or minimized as a function of one or

more variables.4. If Q depends on n variables, where n > 1, find n − 1 equations (constraints)

linking these variables. If this cannot be done the problem cannot be solved bysingle variable techniques.

5. Use the constraints to eliminate the variables and hence express Q as a function ofonly one variable. Determine the interval(s) in which this variable must lie for theproblem to make sense. Alternatively, regard the constraints as implicitly definingn− 1 of the variables, and hence Q, as functions of the remaining variables.

6. Find the required extreme value of the function Q.

15

PURE MATHS IB

7. Make a concluding statement answering the question asked.

Example 2.4.5.A light house L is located on a small island 5 km north of a point A on a straight east-westshoreline. A cable is to be laid from L to a point B on the shoreline 10 km east of A. Thecable will be laid through water on a straight line form L to a point C on the shore linebetween A and B, and from there to B along the shore line. The part of the cable lyingin the water costs $5000/km and the part along the shoreline costs $3000/km

(a) Where should C be located to minimize the cost of the cable?(b) Where should C be located if B is only 3 km from A?

Solution.

(a) Let C be x km from A toward B. Then 0 ≤ x ≤ 10 and LC =√

25 + x2. Thelength CB is 10− x km as illustrated in the figure below

A

L

5 km

xC

√25 + x2

10− x B

Figure 2.9.

Thus the total cost is

T = T (x) = 5000√

25 + x2 + 3000 (10− x) 0 ≤ x ≤ 10.

T is continuous on the closed, finite interval [0, 10], so it attains a minimum valuethat may occur at the endpoints or at a critical point x ∈ (0, 10). To find anycritical points, we set

0 =dT

dx=

5000x√25 + x2

− 3000.

Thus the critical points are at x = ±154 and only x = 15

4 = 3.75 lines in theinterval (0, 10). Now

T (0) = 55000, T (10) = 5000√

125 ≈ 55902, T (3.75) = 50000.

For minimum cost, C should be 3.75 km from A.(b) If B is 3 km, the corresponding total cost function is

T (x) = 5000√

25 + x2 + 3000 (3− x) 0 ≤ x ≤ 3.

This function has the same critical points as in (a) both of which are outside thegiven interval. We thus consider only the endpoints. Therefore T (0) = 34000and T (3) = 29155. To minimize the cost the cable should go straight from L toB.

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PURE MATHS IB

Example 2.4.6.Find the length of the shortest ladder that can extend from a vertical wall, over a fence 2m high located 1 m away from the wall, to a point on the ground outside the fence.

1 m

2 m

L

θ

Figure 2.10.

Solution. Let θ be the angle of inclination of the ladder as shown in the figure above.Then L can be obtained as a function of θ as follows

L (θ) =1

cos θ+

2

sin θ0 < θ <

π

2.

We observe that

limθ→π

2−L (θ) =∞ and lim

θ→0+L (θ) =∞.

Thus L must have a minimum value on(0, π2

)occurring at a critical point. To find

the critical points, set

0 = L′ (θ) = · · · = sin3 θ − 2 cos3 θ

cos2 θ sin2 θ.

At any critical point sin3 θ = 2 cos3 θ ⇒ tan3 θ = 2. Now sec2 θ = 1 + tan2 θ. Butθ = tan−1

(21/3

)so that tan θ = 21/3 and tan2 θ = 22/3. Therefore

sec2 = 1 + 22/3 ⇒ 1

cos2 θ= 1 + 22/3 ⇒ cos θ =

1√1 + 22/3

and

sin θ = cos θ tan θ =21/3√

1 + 22/3.

Therefore the minimum value of L (θ) is

1

cos θ+

2

sin θ=(√

1 + 22/3)1/2

+ 2

(√1 + 22/3

)1/221/3

= · · · ≈ 4.16.

The shortest ladder that can be extended from the wall over the fence to the ground outsideis about 4.16 m long.

Exercise 2.3.

17

PURE MATHS IB

1. Determine whether the given function has any local or absolute extreme values,and find those values if possible

(a) f (x) = x+ 2 on [−1, 1]. Ans: Ab-solute maximum is f (1) = 3 andabsolute minimum is f (−1) = 1.

(b) f (x) = x+ 2 on [−1, 1). Ans: Ab-solute minimum is f (−1) = 1.

(c) f (x) = x2−1 on [−2, 3]. Absoluteminimum value is f (0) = −1 while

absolute maximum is f (3) = 8.Local maximum is f (−2) = 3

(d) f (x) = x3 + x− 4 on [a, b]. Abso-lute maximum is f (b) = b3 + b− 4while absolute minimum is f (a) =a3 + a− 4

(e) f (x) = x2−1. Absolute minimumis f (0) = −1

2. In the following, locate and classify all local extreme values of the given function.Determine whether any of these extreme values are absolute. Sketch the graphof the function.

(a) f (x) = x3 − 3x− 2 .

(b) f (x) = x3 (x− 1)2.

(c) f (x) = x(x2 − 1

)2 (d) f (x) =x2

x2 + 1.

(e) f (x) = x√

2− x2.

3. A billboard is to be maid with 100m2 of printing area and with margins 2m atthe top and bottom and m on each side. Find the outside dimensions of thebillboard if its total area is to be a minimum.

4. A one-meter length of stiff wire is cut into two pieces. One piece is bent intoa circle, the other piece into a square. Find the length of the part used for thesquare if the sum of the areas of the circle and the square is (a) a maximum and(b) a minimum.

5. A window has perimeter 10m and is in the shape of a rectangle with the top edgereplaced with a semicircle. Find the dimensions of the rectangle if the windowadmits the greatest amount of light.

6. Among all rectangles of given area, show that the square has the least perimeter.

2.5. Concavity and Inflections

Definition 2.5.1. We say that a function f is concave up on an open interval Iif it is differentiable there and and if f ′ is an increasing function on I. Similarly, f isconcave down on I if f ′ exists and is decreasing function on I.

Definition 2.5.2. We say that the point (x0, f (x0)) is an inflection point of thecurve y = f (x) if the following two conditions hold

(a) the graph of y = f (x) has a tangent line at x = x0, and(b) the concavity of f is opposite on opposite sides of x0.

Diagram Fig 4.27, 4.28, 4.29

Lemma 2.5.1. Let J be an open interval and let I be an interval consisting of all pointsin J and possibly one or both of the end points of J . Suppose that f is continuous on Iand differentiable on J .

(a) If f ′ (x) > 0 for all x ∈ J , then f is increasing on I.(b) If f ′ (x) < 0 for all x ∈ J , then f is decreasing on I.(c) If f ′ (x) ≥ 0 for all x ∈ J , then f is non-decreasing on I.(d) If f ′ (x) ≤ 0 for all x ∈ J , then f is non-increasing on I.

Theorem 2.7 (Concavity and the second derivative).

(a) If f ′′ (x) > 0 on interval I, then f is concave up on I.(b) If f ′′ (x) < 0 on interval I, then f is concave down on I.

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PURE MATHS IB

+

0

−

−

0

+

a cb x

y

Figure 2.11

(c) If f has an inflection point at x0 and f ′′ (x0) exists, then f ′′ (x0) = 0

Proof.

(a) and (b) follow from applying Lemma 2.5.1 to the derivative f ′ of f .(c) If f has an inflection point at x0 and if f ′′ (x0) exists, then f must be differentiable

in an open interval containing x0. Since f ′ is increasing on one side of x0 anddecreasing on the other side, it must have a local maximum or minimum valueat x0. Thus f ′′ (x0) must be zero.

�

Example 2.5.1.Determine the intervals of concavity of f (x) = x6 − 10x4 and the inflection points of itsgraph.

Solution. We have

f ′ (x) = 6x5 − 40x3, f ′′ (x) = 30x4 − 120x2 = 30x2 (x− 2) (x+ 2)

Thus f ′′ (x) = 0 at x = ±2, 0.Consider the interval (−∞,−2). Here for x = −3,

f ′′ (−3) = 30 (−3)2 (−5) (−1) > 0

and so f is concave up.For x = −1 ∈ (−2, 0), we have

f ′′ (−1) = 30 (−1)2 (−3) (1) < 0

and so f is concave down.For x = 1 ∈ (0, 2), we have

f ′′ (1) = 30 (1)2 (−1) (3) < 0

and so f is concave down.For x = 3 ∈ (2,∞), we have

f ′′ (3) = 30 (3)2 (1) (5) > 0

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PURE MATHS IB

and so f is concave up. We summarize this information in the following chart.

x −2 0 2f ′′ + 0 − 0 − 0 +f ^ infl _ _ infl ^

−3 −2 −1 0 1 2 3

Figure 2.12. f has points of inflection at x = ±2

Example 2.5.2.Determine the intervals of increase and decrease, the local extreme values, the concavityof f (x) = x4 − 2x3 + 1. Use the information to sketch the graph of f .

Solution.

f ′ (x) = 4x3 − 6x2 = 2x2 (2x− 3) , f ′′ (x) = 12x2 − 12x = 12x (x− 1)

Thus f ′′ (x) = 0 when x = 0, 1. f ′ (x) = 0 when x = 0, 1.5For the interval (−∞, 0), take x = −1 and observe that

f ′′ (−1) = 12 (−1) (−2) > 0

so that f is concave up. For the interval (0, 1), take x = 0.5 and obtain

f ′′ (0.5) = 12 (0.5) (−0.5) < 0

so that f is concave up there.

CP CPx 0 1 1.5f ′ - 0 − − 0 +f ′′ + 0 − 0 + +f ↘ infl ↘ infl ↘ min ↗

^ _ ^ ^

Theorem 2.8 (The second derivative test).

(a) If f ′ (x0) = 0 and f ′′ (x0) < 0 then f has a local maximum value at x0.(b) If f ′ (x0) = 0 and f ′′ (x0) > 0 then f has a local minimum value at x0.(c) If f ′ (x0) = 0 and f ′′ (x0) = 0 no conclusion can be drawn; f may have a local

maximum or a local minimum or it may have an inflection point instead.

Proof.

(a) Suppose that f ′ (x0) = 0 and f ′′ (x0) < 0. Since

limh→0

f ′ (x0 + h)

h= lim

h→0

f ′ (x0 + h)− f ′ (x0)h

= f ′′ (x0) < 0

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PURE MATHS IB

Figure 2.13. f has minimum value 4 at x = 2

it follows that f ′ (x0 + h) < 0 for sufficiently small positive h and f ′ (x0 + h) > 0for sufficiently small negative h. By the first derivative test, f must have a localmaximum value at x0.

(b) Exercise(c) No conclusion can be made.

�

Example 2.5.3.Find and classify the critical points of f (x) = x2e−x.

Solution.

f ′ (x) = 2xe−x − x2e−x = xe−x (2− x) = 0 at x = 0 or 2.

f ′′ (x) = · · · =(2− 4x+ x2

)e−x.

Therefore f ′′ (0) = 2 > 0 and f ′′ (2) = −2e−2 < 0. It follows that f has a local minimumvalue at x = 0 and a local maximum value at x = 2.

Exercise 2.4.

Determine the intervals of concavity of the given function, and locate any inflectionpoints.

1. f (x) =√x.

2. f (x) = x2 + 2x+ 3.3. f (x) = 10x3 − 3x5.

4. f (x) =(3− x2

)2.

5. f (x) =(x2 − 4

)3.

2.6. Curve Sketching

When sketching the graph y = f (x) of a function f , we have the following threesources of useful information;

1. the function f itself:- from this we determine the coordinates of some points onthe graph, the symmetry of the graph, any asymptotes and intercepts.

2. the first derivative, f ′:- from this we determine the intervals of increase anddecrease and the location of any local extreme values.

3. the second derivative, f ′′:- from this we determine the the concavity and inflectionpoints, and sometimes the extreme values.

21

PURE MATHS IB

(2, 4e−2

) f (x) = x2e−x

x

y

Figure 2.14. The critical points of f (x) = x2e−x

Asymptotes Asymptotes are straight lines to which a curve draws arbitrarily close as itrecedes to infinite distance from the origin.

Definition 2.6.1. The graph of y = f (x) has a vertical asymptote at x = a if either

limx→a−

f (x) = ±∞ or limx→a+

f (x) = ±∞ or both.

Definition 2.6.2. The graph of y = f (x) has a horizontal asymptote at y = L ifeither

limx→∞

f (x) = L or limx→−∞

f (x) = L or both.

Definition 2.6.3. The straight line y = ax+ b (a 6= 0) is an oblique asymptote of thegraph of y = f (x) if either

limx→∞

(f (x)− (ax+ b)) = 0 or limx→−∞

(f (x)− (ax+ b)) = 0 or both.

Suppose that f (x) = Pm (x) /Qn (x), where Pm and Qn are polynomials of degree mand n respectively. Suppose also that Pm and Qn have no common linear factor. Then

(a) The graph of f has a vertical asymptote at every position x such that Qn (x) = 0.(b) The graph of f has a two-sided horizontal asymptote (i.e. the graph approaches

the asymptotes as x approaches both positive and negative infinity) y = 0 ifm < n.

(c) The graph of f has two-sided horizontal asymptote y = L, (L 6= 0) if m = n. Lis the quotient of the coefficients of the highest terms in Pm and Qn.

(d) The graph of f has two-sided oblique asymptotes if m = n+ 1. This asymptotecan be found by dividing Qn by Pm to obtain a linear quotient, ax + b, and aremainder R, a polynomial of degree at most n− 1, i.e.

f (x) = ax+ b+R (x)

Qn (x).

The oblique asymptote is y = ax+ b.

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PURE MATHS IB

(e) The graph f has no horizontal or oblique asymptote if m > n+ 1.

Example 2.6.1.

Find the oblique asymptote of y = x3

x2+x+1.

Solution. In this case m = 3 = 2 + 1 = n+ 1 and so by long division we obtain

x3

x2 + x+ 1= x− 1 +

1

x2 + x+ 1.

Therefore the oblique asymptote is y = x− 1.

Example 2.6.2.

Sketch the graph of y = x2+2x+42x .

Solution.

y =x2 + 2x+ 4

2x=x

2+ 1 +

2

x.

∴ y′ (x) =1

2− 2

x2and y′′ (x) =

4

x3.

From yDomain: all x 6= 0.Vertical asymptote: x = 0Oblique asymptote: y = x

2 + 1 since y −(x2 + 1

)= 2

x → 0 as x→ ±∞x2 + 2x+ 4 = (x+ 1)2 + 3 ≥ 3 for all x. y is not defined at x = 0.

From y′

Critical points: x2−42x2

= 0 ⇒ x = ±2. The points are (−2,−1) and (2, 3). y′ is not definedat x = 0.From y′′

y′′ is not defined at x = 0 and y′′ = 0 nowhere.Facts about the graph are summarized in the chart below

CP ASY CPx −2 0 2y′ + 0 − undef − 0 +y′′ − 0 undef + +y ↗ max ↘ undef ↘ min ↗

_ _ ^ ^

See Figure 2.15

Example 2.6.3.

Sketch the graph of f (x) = x2−1x2−4 .

Solution.

f ′ (x) = · · · = −6x

(x2 − 4)2and f ′′ (x) = · · · =

6(3x2 + 4

)(x2 − 4)3

.

From fDomain: all x 6= ±2.Vertical asymptotes: x = ±2 where the denominator x2 − 4 = 0

Horizontal asymptotes: limx→±∞

f (x) =1− 0

1− 0= 1

⇒ horizontal asymptote is y = 1Symmetry: is about the y−axis since the function is even.Intercepts: When y = 0, x = ±1, and when x = 0, y = 1

4Hence the intercepts are (0, 1/4), (−1, 0) and (1, 0).

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PURE MATHS IB

y = x2+2x+42x

y = x2 + 1

(2, 3)

(−2,−1)x

y

Figure 2.15.

The two vertical asymptotes divide the graph into three components. The outer com-ponents require points with |x| > 2. WOLG we consider the points (−3, 8/5) and (3, 8/5).From f ′

Critical point is at x = 0, and f ′ is not defined at x = ±2.From f ′′

f ′′ (x) = 0 nowhere and f ′′ not defined at x = ±2.

ASY CP ASYx −2 0 2f ′ − undef + 0 − undef −f ′′ + undef − − undef +f ↗ undef ↗ max ↘ undef ↘

^ _ _ ^

See Figure 2.16

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PURE MATHS IB

y = x2−1x2−4

x=

2x=

2

(3, 8/5)

y = 11

1/4−1

(3−, 8/5)

x

y

Figure 2.16.

Exercise 2.5.

Sketch the graphs of the given functions

1. f (x) =(x2 − 1

)3.

2. f (x) = 2−xx .

3. f (x) = x3

1+x .

4. f (x) = 12−x2 .

5. f (x) = x3

x2+1.

6. In the following, sketch the graph of a function that has the given properties.Identify any critical points, local maxima and minima, and inflection points. As-sume that f is continuous and its derivative exists everywhere unless the contraryis implied or explicitly stated.(a) f (0) = 1, f (±1) = 0, f (2) = 1, limx→∞ f (x) = 2, limx→−∞ f (x) = −1,

f ′ (x) > 0 on (−∞, 0) and on (1,∞), f ′ (x) < 0 on (0, 1), f ′′ (x) > 0 on(−∞, 0) and on (0, 2), and f ′′ (x) < 0 on (2,∞).

(b) f (−1) = 0, f (0) = 2, f (1) = 1, f (2) = 0, f (3) = 1,limx→±∞ (f (x) + 1− x) = 0, f ′ (x) > 0 on (−∞,−1), (−1, 0) and on (2,∞),f ′ (x) < 0 on (0, 2), limx→−1 f

′ (x) = ∞, f ′′ (x) > 0 on (−∞,−1) and on(1, 3), and f ′′ (x) < 0 on (−1, 1) and on (3,∞).

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PURE MATHS IB

2.7. Linear Approximations

2.8. Taylor Polynomials

26

CHAPTER 3

Integration

3.1. Antiderivatives

Definition 3.1.1. An antiderivative of a function f on an interval I is another func-tion F satisfying

F ′ (x) = f (x) for x in I.

Example 3.1.1.

(a) F (x) = x is an antiderivative of the function f (x) = 1 on any interval sinceF ′ (x) = 1 = f (x) everywhere.

(b) G (x) = 12x

2 is an antiderivative of g (x) = x on any interval since G′ (x) = x =g (x) everywhere.

(c) R (x) = −13 cos (3x) is an antiderivative of r (x) = sin (3x) on any interval since

R′ (x) = sin (3x) = r (x) everywhere.(d) F (x) = − 1

x is an antiderivative of f (x) = 1x2

on any interval not containing

x = 0 since F ′ (x) = 1x2

= f (x) everywhere x 6= 0.

We note that the antiderivative of a function is not unique. For suppose F (x) and G (x)are antiderivatives of the function f (x), on an interval I, then

ddx (G (x)− F (x)) = f (x)− f (x) = 0

⇒ G (x)− F (x) = C ⇒ G (x) = F (x) + C on I.

Thus for a given function f (x) on an interval I, different antiderivatives can be obtainedby adding constants to a given antiderivative.

Definition 3.1.2. The indefinite integral of a function f (x) on an interval I is∫f (x)dx = F (x) + C on I,

provided F ′ (x) = f (x) for all x in I. Here C is called the constant of integration.

Example 3.1.2.

(a) since ddx

(12x

2)

= x, then∫xdx =

1

2x2 + C.

(b) since ddx

(14x

4 − 53x

3 + 7x)

= x3 − 5x2 + 7, then∫ (x3 − 5x2 + 7

)dx =

1

4x4 − 5

3x3 + 7x+ C.

(c) since ddx

(− 1x + 4

√x)

= 1x2

+ 2√x

for all x > 0, we have∫ (1

x2+

2√x

)dx = −1

x+ 4√x+ C.

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PURE MATHS IB

(d) since ddx sinx = cosx it follows that∫

(cosx)dx = sinx+ C.

Example 3.1.3.Find the function f (x) whose derivative is f ′ (x) = 6x2 − 1 for all real x and for whichf (2) = 10.

Solution. Since f ′ (x) = 6x2 − 1, then

f (x) =

∫ (6x2 − 1

)dx = 2x3 − x+ C.

Now f (2) = 2 · 23 + C = 10 ⇒ C = −4 so that f (x) = 2x3 − x− 4.

Example 3.1.4.Find the function g (t) whose derivative is (t+ 5) /t3/2 and whose graph passes throughthe point (4, 1).

Solution. We have

g (t) =

∫t+ 5

t32

dt =

∫t−

12 + 5t−

32dt = 2t

12 − 10t−

12 + C.

Now g (4) = 2 · 412 − 10 · 4−

12 + C = 1 implies that C = 2 so that

g (t) = 2t12 − 10t−

12 + 2, t > 0.

Exercise 3.1.

Find the given indefinite integrals

1.

∫5dx.

2.

∫ (2x1/2 + 3x1/3

)dx.

3.

∫x3dx.

4.

∫tanx cosxdx.

5.

∫ (a2 − x2

)dx.

6.

∫ √xdx.

7.

∫1 + cos3 x

cos2 xdx.

3.2. The Definite Integral

In this discussion we assume that the function whose definite integral we define iscontinuous on its domain [a, b]. Let P be the set of points arranged in order between aand b on the real line;

P = {x0, x1, . . . , xn}where a = x0 < x1 < · · · < xn = b. Such a set P is called a partition of [a, b] and itdivides the interval into n subintervals [xi−1, xi]. Note that n depends on P i.e. n = n (P ).The length of the subinterval is

∆xi = xi − xi−1, i = 1, 2, . . . , n.

The greatest of these lengths is called the norm of P and is denoted by ‖P‖, i.e.

‖P‖ = max1≤i≤n

∆xi.

For a point ci ∈ [xi−1, xi], the sum

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PURE MATHS IB

R (f, P, c) =

n∑i=1

f (ci) ∆xi

is called Riemann sum of f on [a, b] corresponding to the partition P and tags C. If thelimit as the number n of the subintervals increases to infinity exists then the function isintegrable on [a, b] and

limn→∞

R (f, P, c) = I =

∫ b

af (x)dx.

We call I the definite integral of f on [a, b]. In this notation

(i)∫

is the integral sign, it resembles the letter S since it represent the limit of asum.

(ii) a and b are the lower and upper limits of integration respectively.(iii) f is called the integrand and x the variable of integration.(iv) dx is the differential of x. It replaces ∆x in the Riemann sums when the limit

is determined.

Note that the definite integral is a number and not a function. The variable x is adummy variable and can be replaced with another variable without affecting the valueof the definite integral.

Example 3.2.1.Express the following limits as definite integrals

(a) limn→∞

n∑i=1

2

n

(1 +

2i− 1

n

)1/3

(b) limn→∞

n∑i=1

1

n

√i

n.

Solution.

(a) We are to interpret the sum as a Riemann sum of f (x) = (1 + x)1/3. From thefactor 2/n we deduce that the length of the interval is 2 and it has been dividedinto n subintervals of equal width ∆xi = 2/n. Let ci = (2i− 1) /n, i = 1, 2, . . . , n.Then

c1 =2− 1

n=

1

n→ 0 as n→∞ and cn =

2n− 1

n=

2− 1/n

1→ 2 as n→∞.

Hence the interval is [0, 2]. The partition is given by xi = 2in , i = 0, 1, . . . , n.

Thus

xi−1 = xi −∆xi =2i

n− 2

n=

2i− 2

n<

2i− 1

n<

2i

n= xi

Thus the sum is indeed a Riemann sum for f (x) over [0, 2]. Since f is continuouson that interval, it is integrable there and

limn→∞

n∑i=1

2

n

(1 +

2i− 1

n

)1/3

=

∫ 2

0(1 + x)1/3dx.

(b) We are to interpret the sum as a Riemann sum of f (x) =√x. From the factor

1/n we deduce that the length of the interval is 1 and it has been divided into nsubintervals of equal width ∆xi = 1/n. Let ci = i/n, i = 1, 2, . . . , n. Then

c1 =1

n→ 0 as n→∞ and cn =

n

n= 1→ 1 as n→∞.

Hence the interval is [0, 1]. The partition is given by xi = in , i = 0, 1, . . . , n. Thus

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PURE MATHS IB

xi−1 = xi −∆xi =i

n− 1

n=i− 1

n<i

n= xi

Thus the sum is indeed a Riemann sum for f (x) over [0, 2]. Since f is continuouson that interval, it is integrable there and

limn→∞

n∑i=1

1

n

√i

n=

∫ 1

0

√xdx.

Properties of the Definite IntegralLet f and g be integrable on an interval containing a, b and c. Then

1.recall thepartition of the

interval

An integral over an interval of zero length is zero.

∫ a

af (x)dx = 0.

2. Reversing the limits of integration changes the sign of the integral

∫ b

af (x)dx = −

∫ a

bf (x)dx.

3. An integral depends linearly on the integrand. If A and B are constants, then

∫ b

a(Af (x) +Bg (x))dx = A

∫ b

af (x)dx+B

∫ b

ag (x)dx.

4. An integral depends additively on the interval of integration.

∫ b

af (x)dx+

∫ c

bf (x)dx =

∫ c

af (x)dx.

5. If a ≤ b and f (x) ≤ g (x) for a ≤ x ≤ b, then

∫ b

af (x)dx ≤

∫ b

ag (x)dx.

6. The triangle inequality for sums extends to definite integrals. If a ≤ a, then

∣∣∣∣∫ b

af (x)dx

∣∣∣∣ ≤ ∫ b

a|f (x)|dx.

7. The integral of an odd function over an interval symmetric about zero is zero. Iff is an odd function (i.e. f (−x) = −f (x)), then

∫ a

−af (x)dx = 0.

8. The integral of an even function over an interval symmetric about zero is twicethe integral over the positive half of the interval. If f is an odd function (i.e.f (−x) = f (x)), then

∫ a

−af (x)dx = 2

∫ a

0f (x)dx.

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PURE MATHS IB

9. There exists a point d ∈ [a, b] such that

∫ b

af (x)dx = (b− a) f (d) .

The value f (d) is called the average value or mean value of f on [a, b].

Exercise 3.2.

1. Calculate L (f, Pn) and U (f, Pn) for the given function f over the given interval [a, b],where Pn is the partition of the interval into n subintervals of equal length ∆x = (b− a) /n.Show that

limn→∞

L (f, Pn) = limn→∞

U (f, Pn) .

(a) f (x) = x on [a, b] = [0, 1].(b) f (x) = 1− x on [a, b] = [0, 2].

(c) f (x) = x3 on [a, b] = [0, 1].(d) f (x) = ex on [a, b] = [0, ].

2. Express the given limit as a definite integral

(a) limn→∞

n∑i=1

π

nsin

(πi

n

)(b) lim

n→∞

n∑i=1

1

n

√i− 1

n

(c) limn→∞

n∑i=1

2

nln

(1 +

2i

n

)

3.3. The Fundamental Theorem of Calculus

Theorem 3.1. Suppose that the function f is continuous on an interval I containingthe point a.

(a) Let the function F be defined on I by

F (x) =

∫ x

af (t)dt.

Then F is differentiable on I and F ′ (x) = f (x) there. Thus F is an antideriva-tive of f on I;

d

dx=

∫ x

af (t)dt = f (x) .

(b) If G (x) is an antiderivative of f on I, so that G′ (x) = f (x) on I, then for anyb in I we have

∫ b

af (x)dx = G (b)−G (a) .

Proof.

(a) Using the definition of the derivative we have

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PURE MATHS IB

F ′ (x) = limh→0

F (x+ h)− F (x)

h

= limh→0

1

h

(∫ x+h

af (t)dt−

∫ x

af (t)dt

)

= limh→0

1

h

(∫ x

af (t)dt+

∫ x+h

xf (t)dt−

∫ x

af (t)dt

)

= limh→0

1

h

(∫ x+h

xf (t)dt

)

= limh→0

(x+ h− x) f (c)

hc ∈ (x, x+ h) by mean value property 9.

= limh→0

h

hf (c)

= limc→x

f (c) ∵ c→ x as h→ 0 = f (x) .

(b) If G′ (x) = f (x), then F (x) = G (x) + C on I for some constant C. Hence

∫ x

af (t)dt = F (x) = G (x) + C.

Let x = a and obtain

0 = G (a) + C ⇒ C = −G (a) .

Next let x = b and obtain

∫ b

af (t)dt = G (b) + C = G (b)−G (a) .

�

To facilitate evaluation of the definite integral using the FTC, we introduce the eval-uation symbol

F (x)

∣∣∣∣ba

= F (b)− F (a) .

Thus

∫ b

af (x)dx =

(∫f (x)dx

) ∣∣∣∣ba

where∫f (x)dx is the indefinite integral of f . Therefore

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PURE MATHS IB

∫ b

af (x)dx =

(∫f (x)dx

) ∣∣∣∣ba

= (F (x) + C)

∣∣∣∣ba

= (F (b) + C)− (F (a) + C)

= F (b)− F (a) = F (x)

∣∣∣∣ba

.

Example 3.3.1.

Evaluate (a)

∫ a

0x2dx (b)

∫ 2

−1

(x2 − 3x+ 2

)dx.

Solution.

(a) ,

∫ a

0x2dx =

1

3x3∣∣∣∣a0

=1

3a3.

(a) ,

∫ 2

−1

(x2 − 3x+ 2

)dx =

(1

3x3 − 3

2x2 + 2x

) ∣∣∣∣2−1

=9

2.

Exercise 3.3.

1. Simplify the following

(a)

∫ b

af (x)dx+

∫ c

bf (x)dx+

∫ c

af (x)dx.

(b)

∫ 2

03f (x)dx+

∫ 3

13f (x)dx−

∫ 3

02f (x)dx−

∫ 2

13f (x)dx.

2. Evaluate the following integrals by using the properties of the definite integraland interpreting integrals as areas.

(a)

∫ 2

−2x+ 2dx.

(b)

∫ b

axdx.

(c)

∫ √2−√2

√2− t2dt.

(d)

∫ π

−πsin(x3)dx.

3. Evaluate the following definite integrals.

(a)

∫ 2

0x3dx.

(b)

∫ 1

1/2

1

x2dx.

(c)

∫ 2

−1

(3x2 − 4x+ 2

)dx.

(d)

∫ 2

−2sin(x2 + 3

)2dx.

4 Find

∫ 2

0f (x) dx, where f (x) =

x2 if 0 ≤ x ≤ 1

x if 1 < x ≤ 2.

33

CHAPTER 4

Methods of Integration

4.1. Method of Substitution

When an integral cannot be solved by inspection. the most important technique is themethod of substitution which we describe below. Recall that by chain rule,

d

dxf (g (x)) = f ′ (g (x)) g′ (x) .

Thus

∫f ′ (g (x)) g′ (x)dx = f (g (x)) + C

and if u = g (x), then du = g′ (x) dx so that

∫f ′ (u)du = f (u) + C.

Example 4.1.1.Find the indefinite integrals

(a)∫

xx2+1

dx

(b)∫ sin(3 lnx)

x dx

(c)∫ex√

1 + exdx.

Solution.

(a) Let u = x2 + 1. Then du = 2xdx. Therefore

∫x

x2 + 1dx =

1

2

∫du

u

=1

2ln |u|+ C =

1

2ln(x2 + 1

)+ C = ln

√x2 + 1 + C.

(b) Let u = 3 lnx. Then du = 3xdx. Therefore

∫sin (3 lnx)

xdx =

1

3

∫sinudu

=1

3(− cosu) + C = −1

3cos (3 lnx) + C.

(c) Let u = 1 + ex. Then du = exdx. Therefore

∫ex√

1 + exdx =

∫u1/2du

=2

3u3/2 + C =

2

3(1 + ex)3/2 + C.

34

PURE MATHS IB

When we introduce the limits of integration then this method is as stated in thetheorem below

Theorem 4.1. Suppose that g is a differentiable function on [a, b] that satisfies g (a) =A and g (b) = B. Also suppose that f is continuous on the range of g. Then

∫ b

af (g (x)) g′ (x)dx =

∫ B

Af (u)du, where u = g (x) .

Example 4.1.2.Evaluate

I =

∫ 8

0

cos√x+ 1√

x+ 1dx.

Solution. Let u =√x+ 1, then du = dx

2√x+1

. If x = 0, u = 1 and if x = 8, u = 3.

Therefore

∫ 8

0

cos√x+ 1√

x+ 1dx = 2

∫ 3

1cosudu = (2 sinu)

∣∣∣∣31

= 2 (sin 3− sin 1) .

ALTERNATIVELY

∫ x=8

x=0

cos√x+ 1√

x+ 1dx = 2

∫ x=8

x=0cosudu = (2 sinu)

∣∣∣∣x=8

x=0

=(2 sin

√x+ 1

) ∣∣∣∣x=8

x=0

= 2 (sin 3− sin 1) .

Example 4.1.3.

Find the area of the region bounded by y =(2 + sin

(x2

))2cos(x2

), the x−axis, and the

lines x = 0 and x = π.

π

y =(2 + sin

(x2

))2cos(x2

)

x

y

Figure 4.1

Solution. Because y ≥ 0 when 0 ≤ x ≤ π, the required area is

A =

∫ π

0

(2 + sin

(x2

))2cos(x

2

)dx.

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PURE MATHS IB

Let u = 2 + sin(x2

), then du = 1

2 cos(x2

)dx. Therefore

A = 2

∫ 3

2u2du =

2

3

(u3) ∣∣∣∣3

2

= · · · = 38

3sq. units.

Example 4.1.4.Evaluate

∫tanxdx.

Solution.∫tanxdx =

∫sinx

cosxdx.

Let u = cosx, then du = −sinxdx. Therefore∫tanxdx =

∫du

u= − ln |u|+ C = − ln |cosx|+ C = ln

∣∣∣∣ 1

cosx

∣∣∣∣+ C.

Exercise 4.1.

Evaluate the following integrals:

1.

∫e5−2xdx

2.

∫ √3x+ 4dx

3.

∫x

(4x2 + 1)5dx

4.

∫xex

2dx

5.

∫cosx

4 + sin2 xdx

6.

∫ex + 1

ex − 1dx

7.

∫x+ 1√

x2 + 2x+ 3dx.

4.2. Integration by Parts

Suppose that U (x) and V (x) are two differentiable functions. By product rule

d

dx(U (x)V (x)) = U (x)

dV

dx+ V (x)

dU

dx.

From this we deduce that∫U (x)

dV

dxdx =

∫d

dx(U (x)V (x))dx−

∫V (x)

dU

dxdx

or simply∫UdV = UV −

∫V dU.

Example 4.2.1.Use integration by parts to evaluate

(a)∫

lnxdx(b)

∫x2 sinxdx

(c)∫x tan−1 xdx

(d)∫

sin−1 xdx

Solution.

(a) Let U = lnx and dV = dx so that dU = 1xdx and V = x. Then∫

lnxdx = x lnx−∫x · 1

xdx = x lnx− x+ C.

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PURE MATHS IB

(b) Let U = x2 and dV = sinxdx so that dU = 2xdx and V = − cosx. Then∫x2 sinxdx = −x2 cosx+

∫cosx (2xdx)

Applying integration by parts to the last integral again, let U = x and dV =cosxdx so that dU = dx and V = sinx. Then∫

(cosx)xdx = x sinx−∫

sinxdx = x sinx+ cosx+ C1.

Therefore∫x2 sinxdx = −x2 cosx+ 2x sinx+ 2 cosx+ C.

(c) Let U = tan−1 x, and dV = xdx. Then tanU = x so that ddx tanU = d

dxx. This

yields sec2 U dUdx = 1 which in turn yields

dU

dx=

1

sec2 U=

1

1 + tan2 U=

1

1 + x2.

From the equation involving dV we obtain V = 12x

2. Therefore

I =

∫x tan−1 xdx =

1

2x2 tan−1 x− 1

2

∫x2

1 + x2dx.

By long division the last integrand can be written as 1− 11+x2

so that

I =1

2x2 tan−1 x− 1

2

∫ (1− 1

1 + x2

)dx =

1

2x2 tan−1 x− 1

2x+

1

2tan−1 x+ C.

(d) Let U = sin−1 x so that sinU = x. Then differentiating both sides of this equationw.r.t. x and simplifying we obtain

cosUdU

dx= 1 ⇒ dU =

dx

cosU=

dx√1− x2

.

On the other dV = dx implying that V = x. Therefore∫sin−1 xdx = x sin−1 x−

∫x√

1− x2.

From the last integral, let U = 1− x2 so that dU = −2xdx. Therefore∫sin−1 xdx = x sin−1 x−1

2

∫u−

12du = x sin−1 x+u

12 +C = x sin−1 x+

√1− x2+C.

Exercise 4.2.

Evaluate the following integrals:

1. ∫x cosxdx

2. ∫x2 cosπxdx

3. ∫x3 lnxdx

4. ∫tan−1 xdx

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PURE MATHS IB

5. ∫x sin−1 xdx

6. ∫ π/4

0sec5 xdx

4.3. Reduction Formulas

Suppose we want to evaluate∫x4e−xdx. For n ≥ 0, let

In =

∫xne−xdx.

We then apply integration by parts as follows. Let U = xn and dV = e−xdx so thatdU = nxn−1 and V = −e−x. Then

In = −xne−x + n

∫xn−1e−xdx = −xne−x + nIn−1.

This is called reduction formula because it gives the value of In in terms In−1, an integralcorresponding to a reduced value of the exponent n.

The above example can now be evaluated as follows

I0 =

∫x0e−xdx =

∫e−xdx = −e−1 + C1.

I1 =

∫xe−xdx = −xe−1 + I0 = −e−1 (x+ 1) + C2.

I2 =

∫x2e−xdx = −x2e−1 + 2I1 = −e−1

(x2 + 2x+ 2

)+ C3.

I3 =

∫x3e−xdx = −x3e−1 + 3I2 = −e−1

(x3 + 3x2 + 6x+ 6

)+ C4.

I4 =

∫x4e−xdx = −x4e−1 + 4I3 = −e−1

(x4 + 4x3 + 12x2 + 24x+ 24

)+ C5.

Example 4.3.1.Obtain and use a reduction formula to evaluate

In =

∫ π/2

0cosn xdx (n = 0, 1, 2, 3, . . .) .

Solution.

I0 =

∫ π/2

0dx = (x)

∣∣∣∣π/20

=π

2and In =

∫ π/2

0cosxdx = (sinx)

∣∣∣∣π/20

= 1.

Now let n ≥ 2 and factorize the integrand as follows

In =

∫ π/2

0cosn xdx =

∫ π/2

0cosn−1 x cosxdx.

Using integration by parts, let U = cosn−1 x and dV = cosxdx. Then dU =(n− 1) cosn−2 x (− sinx) = − (n− 1) cosn−2 x sinx and V = sinx. Therefore

38

PURE MATHS IB

In =(sinx cosn−1 x

) ∣∣∣∣π/20

+ (n− 1)

∫ π/2

0cosn−2 x sin2 xdx

= (n− 1)

∫ π/2

0cosn−2 x

(1− cos2 x

)dx

= (n− 1)

∫ π/2

0cosn−2 xdx− (n− 1)

∫ π/2

0cosn xdx

= (n− 1) In−2 − (n− 1) In.

Solving for In this equation simplifies to

In =n− 1

nIn−2, n ≥ 2.

When n ≥ 2 is even we have

In =n− 1

nIn−2

=n− 1

n

((n− 2)− 1

n− 2In−4

)=

(n− 1) (n− 3)

n (n− 2)In−4

=n− 1

n· n− 3

n− 2

((n− 4)− 1

n− 4In−6

)

=n− 1

n· n− 3

n− 2· n− 5

n− 4In−6

= · · · = n− 1

n· n− 3

n− 2· n− 5

n− 4· · · 5

6· 3

4· 1

2I0

=n− 1

n· n− 3

n− 2· n− 5

n− 4· · · 5

6· 3

4· 1

2· π

2

When n ≥ 2 is odd we have

In = · · · = n− 1

n· n− 3

n− 2· n− 5

n− 4· · · 6

7· 4

5· 2

3I1

=n− 1

n· n− 3

n− 2· n− 5

n− 4· · · 6

7· 4

5· 2

3

Exercise 4.3.

1. Obtain a reduction formula for In =∫

(lnx)dx, and use it to determine I4.2. Obtain a reduction formula for In =

∫sinn xdx (where n ≥ 2), and use it to

determine I6 and I7.3. By writing

In =

∫dx

(x2 + a2)n=

1

a2

∫dx

(x2 + a2)n−1− 1

a2

∫x

x

(x2 + a2)ndx

and integrating the last integral by parts, using U = x, obtain a reduction formulafor In. Use this formula to find I3.

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PURE MATHS IB

4.4. Integrals of Rational Functions

A polynomial is a function P of the form

P (x) = anxn + an−1x

n−1 + · · ·+ a2x2 + a1x+ a0

where n is a nonnegative integer, a0, a1, . . ., an are constants with an 6= 0. We call n thedegree of P .

A quotient P (x) /Q (x) of two polynomials is called a rational function. We gener-ally concern ourselves with cases where degree of P is less than degree of Q. Otherwisethe first thing to do is to simplify the rational function by means of division.

Example 4.4.1.Evaluate

∫x3 + 3x2

x2 + 1dx.

Solution. By long division the integrand simplifies to

x3 + 3x2

x2 + 1= x+ 3− x+ 3

x2 + 1= x+ 3− x

x2 + 1− 3

x2 + 1.

Therefore

∫x3 + 3x2

x2 + 1dx =

∫(x+ 3)dx−

∫x

x2 + 1dx−

∫3

x2 + 1dx

=1

2x2 + 3x− 1

2ln(x2 + 1

)− 3 tan−1 x+ C.

Example 4.4.2.Evaluate∫

x

2x− 1dx.

Solution. The integrand can be simplified by long division as follows

x

2x− 1=

1

2− 1/2

2x− 1=

1

2

(1 +

1

2x− 1

).

Therefore

∫x

2x− 1dx =

1

2

∫ (1 +

1

2x− 1

)dx

=1

2

(x+

1

2ln |2x− 1|

)+ C

=1

2x+

1

4ln |2x− 1|+ C.

Theorem 4.2. Let P and Q be polynomials with real coefficients, and suppose that thedegree of P is less than the degree of Q.

(a) Q (x) can be factored into the product of a constant K, real linear factors of theform x − ai and real quadratic factors of the form x2 + bix + ci having no realroots. The linear and quadratic factors may be repeated.

40

PURE MATHS IB

Q (x) = K (x− a1)m1 (x− a2)m2 · · · (x− aj)mj(x2 + b1x+ c1

)n1 · · ·(x2 + bkx+ ck

)nk .The degree of Q is m1 +m2 + · · ·+mj + 2n1 + · · ·+ 2nk.

(b) The rational function P (x) /Q (x) can be expressed as a sum of partial fractionsas follows

(i) Corresponding to each factor (x− a)m of Q (x) the decomposition containsa sum of fractions of the form

A1

x− a+

A2

(x− a)2+ · · ·+ Am

(x− a)m.

(ii) Corresponding to each factor(x2 + bx+ c

)mof Q (x) the decomposition con-

tains a sum of fractions of the form

B1x+ C1

x2 + bx+ c+

B2x+ C2

(x2 + bx+ c)2+ · · ·+ Bnx+ Cn

(x2 + bx+ c)n.

The constants A1, A2, . . ., Am, B1, B2, . . ., Bn, C1, C2, . . ., Cn can be determinedby adding the fractions in the decomposition and equating the coefficients of thelike powers of x in the numerator of the sum with those in P (x).

The procedure described in the theorem is called decomposition of the rational functionby partial fractions.

Example 4.4.3.Evaluate∫

x+ 4

x2 − 5x+ 6dx.

Solution.

x+ 4

x2 − 5x+ 6=

x+ 4

(x− 2) (x− 3)=

A

x− 2+

B

x− 3=

(A+B)x− (3A+ 2B)

x2 − 5x+ 6.

Equating coefficients of like powers of x in the numerators of the first and last fractionswe have

A + B = 1−3A − 2B = 4

⇒ A = −6, B = 7.

Therefore∫x+ 4

x2 − 5x+ 6dx = −6

∫dx

x− 2+ 7

∫dx

x− 3= −6 ln |x− 2|+ 7 ln |x+ 3|+ C.

Example 4.4.4.

Evaluate I =

∫x3 + 2

x3 − xdx.

Solution. Since the degree of the numerator is not less than the degree of the denom-inator, we begin by simplifying the integrand by division and obtain

x3 + 2

x3 − x= 1 +

x+ 2

x3 − x.

We then decompose the last fraction into partial fractions as demonstrated below.

41

PURE MATHS IB

x+ 2

x3 − x=

x+ 2

x (x− 1) (x+ 1)=A

x+

B

x− 1+

C

x+ 1

=(A+B + C)x2 + (B − C)x−A

x3 − xEquating coefficients of like powers of x in the numerators of the first and last fractionswe have

A + B + C = 0B − C = 1

−A = 2⇒ A = −2, B = 3

2 , C = 12 .

Therefore

I =

∫dx+ 2

∫dx

x+

3

2

∫dx

x− 1+

1

2

∫dx

x+ 1

= x− 2 ln |x|+ 3

2ln |x− 1|+ 1

2ln |x+ 1|

Example 4.4.5.

Evaluate

∫2 + 3x+ x2

x (x2 + 1)dx.

Solution. We decompose the integrand into partial fractions as demonstrated below.

2 + 3x+ x2

x (x2 + 1)=A

x+Bx+ C

x2 + 1=

(A+B)x2 + Cx+A

x (x2 + 1).

Equating coefficients of like powers of x in the numerators of the first and last fractionswe have

A + B = 1C = 3

A = 2⇒ B = −1.

Therefore

∫2 + 3x+ x2

x (x2 + 1)dx = 2

∫dx

x−∫

x

x2 + 1dx+ 3

∫dx

x2 + 1

= 2 ln |x| − 1

2ln(x2 + 1

)+ 3 tan−1 x+ C.

Example 4.4.6 (Completing squares).

Evaluate I =

∫dx

x3 + 1.

Solution. Q (x) = x3 + 1 = (x+ 1)(x2 − x+ 1

). Thus the integrand is decomposed

as follows

1

x3 + 1=

A

x+ 1+

Bx+ C

x2 − x+ 1=

(A+B)x2 + (−A+B + C)x+A+ C

x3 + 1.

42

PURE MATHS IB

Equating coefficients of like powers of x in the numerators of the first and last fractionswe have

A + B = 0−A + B + C = 0A C = 1

⇒ A = 13 , B = −1

3 , C = 23 .

Therefore

I =1

3

∫dx

x+ 1− 1

3

∫x− 2

x2 − x+ 1dx

=

[x2 − x+ 1 =

(x− 1

2

)2+ 3

4x− 2 =

(x− 1

2

)− 3

2

]=

1

3ln |x+ 1| −

∫ (x− 1

2

)− 3

2(x− 1

2

)2+ 3

4

dx

=

[let u = x− 1

2∴ du = dx

]=

1

3ln |x+ 1| −

∫u− 3/2

u2 + 3/4du

=1

3ln |x+ 1| − 1

3

(∫u

u2 + 3/4du− 3

2

∫du

u2 + 3/4du

)

=1

3ln |x+ 1| − 1

3

∫u

u2 + 3/4du+

1

2

1√3/2

tan−1(

u√3/2

)

=

[let v = u2 + 3

4∴ dv

2 = udu

]=

1

3ln |x+ 1| − 1

6

∫dv

v+

1√3

tan−1(

2 (x− 1/2)√3

)

=1

3ln |x+ 1| − 1

6ln |v|+ 1√

3tan−1

(2 (x− 1/2)√

3

)+ C

=1

3ln |x+ 1| − 1

6ln

(u2 +

3

4

)+

1√3

tan−1(

2 (x− 1/2)√3

)+ C

=1

3ln |x+ 1| − 1

6ln

((x− 1

2

)2

+3

4

)+

1√3

tan−1(

2 (x− 1/2)√3

)+ C

=1

3ln |x+ 1| − 1

6ln(x2 − x+ 1

)+

1√3

tan−1(

2 (x− 1/2)√3

)+ C

Example 4.4.7 (Denominator with repeated factors).

Evaluate

∫dx

x (x− 1)2.

Solution. The integrand is decomposed as follows

1

x (x− 1)2=A

x+

B

x− 1+

C

(x− 1)2=

(A+B)x2 + (−2A−B + C)x+A

x (x− 1)2.

Equating coefficients of like powers of x in the numerators of the first and last fractionswe have

A + B = 0−2A − B + C = 0A = 1

⇒ B = −1, C = 1.

43

PURE MATHS IB

∫dx

x (x− 1)2=

∫dx

x−∫

dx

x− 1dx+

∫dx

(x− 1)2

= ln |x| − ln |x− 1| − 1

x− 1+ C.

Example 4.4.8.

Evaluate I =

∫x2 + 2

4x5 + 4x3 + x.

Solution. The denominator can be factorised as x(4x4 + 4x2 + 1

)= x

(2x2 + 1

)2.

The integrand is then decomposed as follows

x2 + 2

4x5 + 4x3 + x=A

x+Bx+ C

2x2 + 1+

Dx+ E

(2x2 + 1)2= · · ·

=(4A+ 2B)x4 + 2Cx3 + (4A+B +D)x2 + (C + E)x+A

4x5 + 4x3 + x.

Equating coefficients of like powers of x in the numerators of the first and last fractionswe have

4A + 2B = 02C = 0

4A + B + = 1C + E = 0

A = 2

⇒ A = 2 B = −4 C = 0D = −3 E = 0

I = 2

∫dx

x− 4

∫x

2x2 + 1dx−

∫x

(2x2 + 1)2dx

=

[let u = 2x2 + 1∴ du = 4xdx

]= 2 ln |x| −

∫du

u− 3

4

∫du

u2

= 2 ln |x| − ln |u|+ 3

4

1

u+ C

= ln(x2)− ln

(2x2 + 1

)+

3

4

1

2x2 + 1+ C

= ln

(x2

2x2 + 1

)+

3

4

1

2x2 + 1+ C

Exercise 4.4.

1. Evaluate the following integrals:

(a)

∫2dx

2x− 3

(b)

∫xdx

πx+ 2

(c)

∫1

x2 − 9dx

(d)

∫dx

a2 − x2

(e)

∫x2dx

x2 + x− 2

(f)

∫x− 2

x2 + xdx

44

PURE MATHS IB

2. (a) Writex5

(x2 − 4) (x+ 2)2as the sum of a polynomial and a partial fraction

decomposition.

(b) Hence evaluate

∫x5

(x2 − 4) (x+ 2)2dx.

4.5. Inverse Substitution

4.5.1. The Inverse Trigonometric Substitution. For the substitutions

x = a sin θ, x = a tan θ, and x = a sec θ,

we obtain

θ = sin−1x

a, θ = tan−1

x

a, and θ = sin−1

x

a

respectively.The inverse sine substitution

Integrals involving√a2 − x2 (a > 0) can be reduced to a simpler form by means of the

substitution

x = a sin θ.

Since a2 − x2 > 0 only for −a ≤ x ≤ a the corresponding range for θ is −π/2 ≤ θ ≤ π/2.Since cos θ ≥ 0 for such θ, we have

√a2 − x2 =

√a2 − a2 sin2 θ = · · · = a cos θ.

It follows that

cos θ =

√a2 − x2a

and tan θ =x√

a2 − x2.

Example 4.5.1.

Evaluate

∫dx

(5− x2)3/2.

Solution. Let x =√

5 sin θ. Then dx =√

5 cos θdθ and 5− x2 = 5 cos2 θ. Therefore

∫dx

(5− x2)3/2=

∫ √5 cos θ

(5 cos2 θ)3/2dθ

=1

5

∫cos θ

cos3 θdθ =

1

5

∫dθ

cos2 θ

=1

5

(− 1

cos θ

)(− sin θ) + C

=1

5tan θ + C =

1

5

x√5− x2

+ C.

Example 4.5.2.Find the area of the circular segment shaded in Fig 4.2

45

PURE MATHS IB

y =√a2 − x2

b ax

y

Figure 4.2.

Solution. The area is

A = 2

∫ b

a

√a2 − x2dx Let x = a sin θ

then dx = a cos θdθ and a2 − x2 = a2 cos2 θ

= 2

∫ x=a

x=ba2 cos2 θdθ

= a2∫ x=a

x=b(1 + cos 2θ)dθ ∵ sin2 θ =

1− cos 2θ

2

= a2(θ +

1

2sin 2θ

) ∣∣∣∣x=ax=b

= a2(θ +

1

2· 2 sin θ cos θ

) ∣∣∣∣x=ax=b

= a2 (θ + sin θ cos θ)

∣∣∣∣x=ax=b

= a2

(sin−1

x

a+x

a·√a2 − x2a

)∣∣∣∣x=ax=b

= · · · = a2(π

2− sin−1

b

a

)− b√a2 − b2 square units

The inverse tangent substitutionIntegrals involving

√a2 + x2 or 1

a2+x2are simplified by the substitution

x = a tan θ.

It follows that θ = tan−1 xa and that x can take any real value. Also θ ∈ (−π/2, π/2) sothat sec θ > 0. Therefore

46

PURE MATHS IB

a2 + x2 = a2(1 + tan2 θ

)= a2 sec2 θ, sin θ =

x√a2 + x2

, cos θ =a√

a2 + x2.

Example 4.5.3.

Evaluate (a)

∫dx√

4 + x2and (b)

∫dx

(1 + 9x2)2.

Solution.

(a) Let x = 2 tan θ so that dx = 2 sec2 θdθ and√

4 + x2 = 2 sec θ. Therefore

∫dx√

4 + x2=

∫2 sec2 θ

2 sec θdθ

=

∫sec θdθ

=

∫sec θ (sec θ + tan θ)

sec θ + tan θdθ Let u = sec θ + tan θ

then du = sec θ (sec θ + tan θ) dθ

=

∫du

u= ln |u|+ C = ln |sec θ + tan θ|+ C

= ln

∣∣∣∣∣√

4 + x2

2+x

2

∣∣∣∣∣+ C

(b) Let 3x = tan θ so that 3dx = 2 sec2 θdθ and 1 + 9x2 = sec2 θ. We also have that

cos θ =1√

1 + 9x2and sin θ =

3x√1 + 9x2

.

Therefore

∫dx

(1 + 9x2)2=

1

3

∫sec2 θ

sec4 θdθ

=1

3

∫cos2 θdθ =

1

6

∫(1 + cos 2θ)dθ

=1

6

(θ +

1

2sin 2θ

)+ C

=1

6

(θ +

1

2· 2 sin θ cos θ

)+ C

=1

6tan−1 (3x) +

1

6

3x√1 + 9x2

· 1√1 + 9x2

+ C

=1

6tan−1 (3x) +

1

2

x

1 + 9x2+ C

The inverse secant substitutionIntegrals involving

√x2 − a2 (a > 0) can be simplified by using the substitution

47

PURE MATHS IB

x = a sec θ so that θ = sec−1x

a.

The expression√x2 − a2 only holds for x ≥ a or x ≤ −a.

If x ≥ a, then 0 < ax ≤ 1. Now

θ = sec−1x

a= cos−1

a

x⇒ cos θ =

a

x⇒ 0 < cos θ ≤ 1.

Therefore 0 ≤ θ < π2 so that tan θ ≥ 0 and

√x2 − a2 = a tan θ.

If x ≤ −a, then −1 ≤ ax < 0. Thus

−1 ≤ cos θ < 0 ⇒ π

2< θ ≤ π.

∴ tan θ ≤ 0 and√x2 − a2 = −a tan θ.

Combining these two cases it follows that the

√x2 − a2 = a |tan θ| .

Example 4.5.4.

Evaluate I =

∫dx√x2 − a2

where a > 0.

Solution. Assume x ≥ a. Let x = sec θ so that dx = a sec θ tan θdθ and√x2 − a2 =

a tan θ. Therefore

I =

∫a sec θ tan θ

a tan θdθ =

∫sec θdθ

= ln |sec θ + tan θ|+ C1

= ln

∣∣∣∣∣xa +

√x2 − a2a

∣∣∣∣∣+ C1 = ln

∣∣∣∣∣x+√x2 − a2a

∣∣∣∣∣+ C1

= ln∣∣∣x+

√x2 − a2

∣∣∣+ C2.

If x ≤ −a, let u = −x so that u ≥ a and du = −dx. Therefore

48

PURE MATHS IB

I = −∫

du√u2 − a2

= − ln∣∣∣u+

√u2 − a2

∣∣∣+ C3

= ln

∣∣∣∣ 1

u+√u2 − a2

∣∣∣∣+ C3

= ln

∣∣∣∣ 1

−x+√x2 − a2

∣∣∣∣+ C3

= ln

∣∣∣∣∣ 1

−x+√x2 − a2

· x+√x2 − a2

x+√x2 − a2

∣∣∣∣∣+ C3

= ln

∣∣∣∣∣x+√x2 − a2−a2

∣∣∣∣∣= ln

∣∣∣x+√x2 − a2

∣∣∣+ C4

Thus in either case

I = ln∣∣∣x+

√x2 − a2

∣∣∣+ C.

Exercise 4.5.

Evaluate the following integrals:

1.

∫dx√

1− 4x2

2.

∫x3dx√9 + x2

3.

∫x2dx

(a2 − x2)3/2dx

4.

∫dx

(4x2 + 4x+ 5)2

5.

∫xdx√

2ax− x2

6.

∫ 0

− ln 2ex√

1− e2x dx

7.

∫x2 dx√9− x2

.

4.6. Improper Integrals

Consider the definite integral

I =

∫ b

af (x) dx.

Improper integrals occur under the following cases

(i) Improper integral type I: Here we have a = −∞ or b =∞ or both.(ii) Improper integral type II: Here f is unbounded as x approaches a or b or both.

Definition 4.6.1 (Improper integrals of type I).If f is continuous on [a,∞), then∫ ∞

af (x) dx = lim

R→∞

∫ R

af (x) dx.

Similarly if f is continuous on (−∞, b], then∫ b

−∞f (x) dx = lim

R→−∞

∫ b

Rf (x) dx.

49

PURE MATHS IB

In either case, if the limit exists (is a finite number), we say that the improper integralconverges. If the limit does not exist, we say that the improper integral diverges. If thelimit is ∞ (or −∞), we say that the improper integral diverges to infinity (or negativeinfinity).

Example 4.6.1.

Evaluate

∫ ∞−∞

1

1 + x2dx.

Solution. Fig 6.10

Figure 4.3.

By the symmetry of the integrand we have∫ ∞−∞

1

1 + x2dx =

∫ 0

−∞

1

1 + x2dx+

∫ ∞0

1

1 + x2dx

= 2 limR→∞

∫ R

0

1

1 + x2dx

= 2 limR→∞

tan−1R = 2(π

2

)= π.

Example 4.6.2.∫ ∞0

cosx dx = limR→∞

∫ R

0cosx dx = lim

R→∞sinR.

Fig 6.11

As R increases, the integral alternately adds and subtracts the areas of the hills and thevalleys but does not approach any unique limit.

Definition 4.6.2 (Improper integrals f type II).

50

PURE MATHS IB

If f is continuous on (a, b] and possibly unbounded near a, we have∫ b

af (x) dx = lim

c→c+

∫ c

cf (x) dx.

Similarly if f is continuous on [a, b) and is possibly unbounded near b, we have∫ b

af (x) dx = lim

c→b−

∫ c

af (x) dx.

These improper integral may converge, diverge, diverge to infinity (or negative infinity).

Example 4.6.3.Find the area of the region S lying under y = 1/

√x, above the x−axis, between x = 0

and x = 1.

Solution. Fig 6.12

The area is given by

A =

∫ 1

0

1√xdx,

which is an improper integral of type II since the integrand is unbounded near x = 0, whichis a vertical aymptote. Therefore

A = limc→0+

∫ 1

cx−1/2 dx = lim

c→0+2x1/2

∣∣∣∣1c

= limc→0+

(2− 2

√c)

= 2.

This integral converges, and S has a finite area of 2 square units.

Example 4.6.4.Evaluate each of the following integrals or show that it diverges:

(a)

∫ 1

0

1

xdx, (b)

∫ 2

0

1√2x− x2

dx, (c)

∫ 1

0lnx dx.

Solution.

(a) ∞, (b) π, (c) −1.

Theorem 4.3 (p-integrals). If 0 < a <∞, then

(a)

∫ ∞a

x−p dx

converges toa1−p

p− 1if p > 1

diverges to ∞ if p ≤ 1

(b)

∫ a

0x−p dx

converges toa1−p

1− pif p < 1

diverges to ∞ if p ≥ 1.

Proof.

51

PURE MATHS IB

(a) Here we have∫ ∞a

x−p dx = limR→∞

∫ R

ax−p dx = lim

R→∞

(x−p+1

−p+ 1

) ∣∣∣∣Ra

=1

−p+ 1

(limR→∞

1

Rp−1− a−p+1

)

=

converges to a1−p

p−1 if p > 1

diverges to ∞ if p ≤ 1.

(b) Here we have∫ a

0x−p dx = lim

c→0+

∫ a

cx−p dx = lim

c→0+

(x−p+1

−p+ 1

) ∣∣∣∣ac

=1

−p+ 1

(a−p+1 − lim

c→0+c1−p

)

=

converges to a1−p

1−p if p < 1

diverges to ∞ if p ≥ 1.

�

Problem 1. Show that

∫ ∞0

x−p dx does not converge for any value of p.

Example 4.6.5.

Evaluate

∫ 2

0f (x) dx where f (x) =

1/√x if 0 < x ≤ 1

x− 1 if 1 < x ≤ 2.

Solution. Fig 6.13

∫ 2

0f (x) dx =

∫ 1

0

dx√xdx+

∫ 2

1(x− 1) dx

= limc→0+

∫ 1

0

dx√xdx+

(x2

2− x) ∣∣∣∣2

1

= 2 +

(2− 2− 1

2+ 1

)=

5

2.

In the event that an improper integral cannot be evaluated, its convergence can bedetermined by comparing it with simpler integrals.

52

PURE MATHS IB

Theorem 4.4 (A comparison theorem for integrals). Let −∞ ≤ a < b ≤ ∞,and suppose that the functions f and g are continuous on the interval (a, b) and satisfy

0 ≤ f (x) ≤ g (x). If

∫ b

ag (x) dx converges, then so does

∫ b

af (x) dx, and∫ b

af (x) dx ≤

∫ b

ag (x) dx.

Equivalently, if

∫ b

af (x) dx diverges to ∞, then so does

∫ b

ag (x) dx.

Proof. Since 0 ≤ f (x) ≤ g (x), each integral either converge to a nonnegative numberor diverge to ∞. Since f (x) ≤ g (x) on (a, b), we have that for a < r < s < b, we have∫ s

rf (x) dx ≤

∫ s

rg (x) dx.

The theorem then follows by taking limits as r → a+ and s→ b−. �

Example 4.6.6.

Show that

∫ ∞0

e−x2dx converges, and find an upper bound for its value.

Solution. AS OF NOW it is not possible to integrate e−x2

so we seek function(s)

that can be compared with e−x2

but whose integral(s) can be determined. On the interval

[1,∞), x2 ≥ x, so −x2 ≤ −x and 0 < e−x2 ≤ e−x. On [0, 1] we have 0 < e−x

2 ≤ 1.Therefore

0 <

∫ ∞0

e−x2dx ≤

∫ 1

01 dx+

∫ ∞1

e−x dx

= 1 + limR→∞

e−x

−1

∣∣∣∣R1

= 1 + limR→∞

(1

e− 1

eR

)= 1 +

1

e.

Hence

∫ ∞0

e−x2dx converges and its value is not greater than 1 + (1/e).

Example 4.6.7.Show that if p > 1 and f is continuous on [1,∞) and satisfies the condition |f (x)| ≤

K |x−p| for some constant K for all x ≥ 1, then

∫ ∞1

f (x) dx converges.

Solution.∫ ∞1

f (x) dx = limR→∞

∫ R

1f (x) dx ≤ lim

R→∞

∫ R

1Kx−p dx.

This integral converges toK

p− 1according to part (a) of Theorem 4.3.

Exercise 4.6.

In the following, evaluate the given integral or show that it diverges.

1.

∫ ∞2

1

(x− 1)3dx

2.

∫ ∞3

1

(2x− 1)2/3dx

3.

∫ ∞0

e−2x dx

4.

∫ −1−∞

dx

x2 + 1

53

PURE MATHS IB

5.

∫ 1

−1

dx

(x+ 1)/3

6.

∫ a

0

dx

a2 − x2dx

7

∫ 1

0lnx dx.

State whether the given integral converges or diverges, and justify your claim.

7.

∫ ∞0

x2

x5 + 1dx

8.

∫ ∞0

dx

1 +√x

9.

∫ ∞2

x√x

x2 − 1dx

10.

∫ ∞0

e−x3dx

11.

∫ ∞0

dx√x+ x2

dx

12.

∫ 1

−1

ex

x+ 1dx

54

CHAPTER 5

Applications of Integration

5.1. Areas of Plane Figures

Recall: The definite integral∫ ba f (x)dx measures the area between the graph and the

x−axis from x = a to x = b. However it yields negative values for any part that lies belowthe x−axis.

Fig 5.27

∫ b

af (x)dx = A1 −A2.

To determine the total area as a positive value we instead integrate |f (x)|;

∫ b

a|f (x)|dx = A1 +A2.

Note that

|f (x)| =

f (x) if f (x) > 0

−f (x) if f (x) < 0.

In determining the integrals, we break the domain into subintervals to corresponding topositive and negative values of f (x).

Example 5.1.1.Determine the area bounded by y = cosx, y = 0, x = 0, and x = 3

2π.

Solution. Fig 5.28

Figure 5.1. [-7mm]

55

PURE MATHS IB

A =

∫ 32π

0|cosx|dx

=

∫ π2

0cosxdx+

∫ 32π

π2

− cosxdx

= (sinx)

∣∣∣∣π20

− (sinx)

∣∣∣∣ 32ππ2

= · · · = 3 square units

Suppose the region R is bounded by the graphs of two continuous functions y = f (x)and y = g (x), and the vertical lines x = a and x = b. Assuming that a < b, then the areaof R is obtained by the formula

A =

∫ b

a|f (x)− g (x)|dx.

To evaluate the integral, we have to determine the intervals in which f (x) > g (x) andbreak the integral into a sum of integrals over each of these integrals where the integrandis positive.

Example 5.1.2.Find the area of the bounded plane region R lying between the curves y = x2 − 2x andy = 4− x2.

Solution. We determine the intersection of the curves by solving the equation x2 −2x = 4− x2. The solutions are x = 2 or x = −1.

Fig 5.31

In the entire interval x2 − 2x ≤ 4− x2. Thus

A =

∫ 2

−1

((4− x2

)−(x2 − 2x

))dx

=

∫ 2

−1

(4 + 2x− 2x2

)dx = · · · = 9 square units

56

PURE MATHS IB

Example 5.1.3.Find the total area A lying between the curves y = sinx and y = cosx from x = 0 tox = 2π.

Solution. Points in [0, 2π] where cosx = sinx are x = π/4 + nπ n = 0, 1In the intervals

(0, π4

)and

(5π4 , 2π

), cosx > sinx.

In the interval(π4 ,

5π4

), cosx < sinx.

Fig 5.32

∴ A =

∫ π/4

0(cosx− sinx)dx+

∫ 5π/4

π/4(sinx− cosx)dx

+

∫ 2π

5π/4(cosx− sinx)dx

= (sinx− cosx)

∣∣∣∣π/40

+ (− cosx− sinx)

∣∣∣∣5π/4π/4

+ (sinx− cosx)

∣∣∣∣2π5π/4

= · · · = 4√

2 square units

Exercise 5.1.

Sketch and find the area of the plane region bounded by the given curves

1. y = x, y = x2

2. y = x2 − 5, y = 3− x23. 2y = 4x− x2, 2y + 3x = 6

4. y = x3, y = x5. y = x3, y2 = x6. y = 1

x , 2x+ 2y = 5

7. Find the area of the region above y = x2 − 4x and below the x−axis.

5.2. Volume

Volumes by slicingSuppose that a solid S lies between planes perpendicular to the x−axis at positions x = aand x = b and that the cross-sectional area of S in the plane perpendicular to the x−axisat x is a known function of A (x), for a ≤ x ≤ b. We assume that A (x) is a continuousfunction on [a, b]. If a = x0 < x1 < · · · < xn−1 < xn = b, then P = {x0, x1, . . . , xn−1, xn}is a partition of [a, b] into n sub-intervals, and the perpendicular planes to the x−axis atx1, x2, . . ., xn−1 divide the solid into n slices of which the ith has thickness ∆xi = xi−xi−1.The volume ∆Vi of that slice lies between the maximum and minimum values of A (x) ∆Vifor some x ∈ [xi−1, xi].

Fig 7.2, 7.3

57

PURE MATHS IB

So

∆Vi = A (ci) ∆xi

for some ci ∈ [xi−1, xi]. The volume of the solid is then approximated by the Riemannsum

V =

n∑i=1

∆Vi =

n∑i=1

A (ci) ∆xi.

Letting n→∞ so that ∆xi → 0 we obtain the volume as

V =

∫ b

aA (x)dx

as the limit of the corresponding Riemann sum. This volume can also be given by

V =

∫ b

adV, where dV = A (x)

Solids of revolutionIf a solid has a circular cross-section in a plane perpendicular to some axis, then such asolid is called a solid of revolution since it can be generated by rotating a plane regionabout an axis in that plane so that it sweeps out the solid.

If the region R bounded by y = f (x), y = 0, x = a and x = b is rotated aboutthe x−axis then the cross-section of the solid generated in the plane perpendicular tothe x−axis at x is a circular disk of radius |f (x)|. The area of this cross-section is

A (x) = π (f (x))2 so that the volume of the solid of revolution is

V = π

∫ x=b

x=a(f (x))2dx.

Example 5.2.1 (Volume of a ball).

Find the volume of a solid ball having radius a.

Solution. The ball can be generated by rotating the half disk 0 ≤ y ≤√a2 − x2,

−a ≤ x ≤ a about the x−axisFig 7.5 (a)

Therefore the volume is

V = π

∫ a

−a

(√a2 − x2

)2dx = 2π

∫ a

0

(a2 − x2

)dx

= 2π

(a2x− x3

3

) ∣∣∣∣a0

=4

3πa3 square units

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PURE MATHS IB

Example 5.2.2.Find the volume of a right circular cone of base radius r and height h that is generatedby rotating the triangle with vertices (0, 0), (h, 0) and (h, r) about the x− axis.

Solution. Fig 7.5 (b)

The line from (0, 0) to (h, r) has equation y = rhx. Thus the volume of the cone is

V = π

∫ h

0

(rxh

)2dx = π

( rh

)2(x33

) ∣∣∣∣h0

=1

3πr2h cubic units.

Example 5.2.3.A ring-shaped solid is generated by rotating the finite plane region R bounded by thecurve y = x2 and the line y = 1 about the line y = 2. Find its volume.

Solution. Fig 7.7

Solving the pair of equations y = x2 and y = 1, reveals that the curves intersect atx = ±1. Thus the solid lies between these two values of x.

The area element of R at x is a vertical strip of width dx extending from y = x2 toy = 1. When R is rotated about the line y = 2, this area element sweeps out a thin,washer-shaped volume element of thickness dx and radius 2−x2, having a hole of radius 1through the middle. The cross-sectional area of this element is the difference in the areasof the circles having the same center and radii 2− x2 and 1. Thus

dV =(π(2− x2

)2 − π (1)2)dx = π

(3− 4x2 + x4

)dx.

The solid of revolution is given by

V = π

∫ 1

−1

(3− 4x2 + x4

)dx

= 2π

∫ 1

0

(3− 4x2 + x4

)dx = · · · = 56

15π cubic units.

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PURE MATHS IB

Exercise 5.2.

Find the volume of each solid S in two ways, using the method of slicing and the methodof cylindrical shells

1. S is generated by rotating about the x−axis the region bounded by y = x2, y = 0,and x = 1.

2. S is generated by rotating about the x−axis the region bounded by y = x2 andy =√x between x = 0 and x = 1.

Find the volumes of the solids obtained if the plane regions R described below are rotatedabout (a) the x−axis and (b) the y−axis.

3. R is bounded by y = x (2− x) and y = 0 between x = 0 and x = 2.5. R is finite region bounded by y = x and x = 4y − y2.7. R is the region bounded by y = 1/

(1 + x2

), y = 2, x = 0, and x = 1.

5.3. Arc Length

If A and B are two points in the plane, denote by |AB| the length of the straight linesegment AB. Let C be a curve joining A and B. Choose points A = P0, P1, . . . , Pn−1, Pn =B in order along the curve C as shown below.

Fig 7.21

The polygonal line P0P1 · · ·Pn−1Pn constructed by joining adjacent pairs of these pointswith straight line segments form a polygonal approximation to C , having the length

Ln = |P0P1|+ |P1P2|+ · · ·+ |Pn−1Pn| =n∑i=1

|Pi−1Pi|.

It is easy to see that Ln can never exceed the length of C . It is also easy to see that theapproximation Ln improves as n increases.

Let f be a function defined on a closed interval [a, b] and having a continuous derivativef ′ there. If C is the graph of f , then any partition of the form

P = {a = x0, x1, x2, . . . , xn−1, xn = b}with a = x0 < x1 < x2 < . . . < xn−1 < xn = b induces a polygonal approximation of C .For this partition let Pi be the point (xi, f (xi)), (0 ≤ i ≤ n). Then the length

Ln =

n∑i=1

|Pi−1Pi| =n∑i=1

√(xi − xi−1)2 + (f (xi)− f (xi−1))

2

=

n∑i=1

√1 +

(f (xi)− f (xi−1)

xi − xi−1

)2

∆xi

where ∆xi = xi − xi−1. By MVT ∃ a number ci ∈ [xi−1, xi] such that

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PURE MATHS IB

f (xi)− f (xi−1)

xi − xi−1= f ′ (ci)

so that

Ln =

n∑i=1

√1 + (f ′ (ci))

2 ∆xi

which is a Riemann sum.Thus the arc length s of the curve y = f (x) from x = a to x = b is given by

s =

∫ b

a

√1 + (f ′ (x))2dx =

∫ b

a

√1 +

(dy

dx

)2

dx.

You can regard the arc length defined above as the sum of arc length elements

s =

∫ b

ads, where ds =

√1 + (f ′ (x))2 dx.

Consider the figure belowFig 7.22

The differential triangle suggests that

(ds)2 = (dx)2 + (dy)2(ds

dx

)2

= 1 +

(dy

dx

)2

ds

dx=

√1 +

(dy

dx

)2

ds =

√1 +

(dy

dx

)2

dx =

√1 + (f ′ (x))2 dx

A similar argument shows that for a curve specified by an equation of the form x = g (y),(c ≤ y ≤ d), the arc length element is

ds =

√1 +

(dx

dy

)2

dy =

√1 + (g′ (y))2 dy.

Example 5.3.1.Find the arc length of the curve y = x2/3 from x = 1 to x = 8.

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PURE MATHS IB

Solution. Ans: 40√40−13

√13

27

Example 5.3.2.Find the arc length of the curve y = x4 + 1

32x2from x = 1 to x = 2.

Solution. Here

dy

dx= 4x3 − 1

16x3and 1 +

(dy

dx

)2

= · · · =(

4x3 +1

16x3

)2

.

The expression in the last bracket is positive for x ∈ [1, 2] and so

s =

∫ 2

1

(4x3 − 1

16x3

)dx =

(x4 − 1

32x2

) ∣∣∣∣21

= · · · = 15 +3

128.

Example 5.3.3.Find the circumference of the ellipse

x2

a2+y2

b2= 1

where a ≥ b > 0.

Solution. Fig 7.25

The upper half of the ellipse has the equation

y = b

√1− x2

a2=b

a

√a2 − x2.

∴dy

dx= · · · = − b

a

x√a2 − x2

and so

1 +

(dy

dx

)2

= · · · =a4 −

(a2 − b2

)x2

a2 (a2 − x2).

The circumference of the ellipse is four times the arc length of the part lying in thefirst quadrant, so

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PURE MATHS IB

s = 4

∫ a

0

√a4 − (a2 − b2)x2

a√a2 − x2

dx

=

[x = a sin t

dx = a cos tdt

]= 4

∫ π/2

0

√a4 − (a2 − b2) a2 sin2 t

a√a2 − a2 sin2 t

a cos t dt

= 4

∫ π/2

0

√a4 − (a2 − b2) sin2 t dt

= 4a

∫ π/2

0

√1− a2 − b2

a2sin2 t dt

= 4a

∫ π/2

0

√1− ε2 sin2 t dt

where

ε =

√a2 − b2a

is the eccentricity of the ellipse.The function E (ε), defined by

E (ε) =

∫ π/2

0

√1− ε2 sin2 tdt

is called the complete elliptic integral of the second kind. The integral cannot beevaluated analytically for general ε; the alternative is numerical integration.

Exercise 5.3.

Find the lengths of the given curves

1. y = 2x− 1 from x = 1 to x = 3.2. y = 2

3x3/2 from x = 0 to x = 8.

3. y3 = x2 from (−1, 1) to (1, 1).

4. y = x3

12 + 1x from x = 1 to x = 4.

5. 4y = 2 lnx − x2 from x = 1 tox = e.

6. y = ex+e−x

2 (= coshx) from x = 0to x = a.

5.4. Areas of Surfaces of Revolution

When a plane curve is rotated about a line in the plane of the curve, it sweeps out asurface of revolution. If the radius of rotation of the arc length element ds is r, then itgenerates, on rotation, a circular band of width ds and length (circumference) 2πr. Thearea of this band is therefore

dS = 2πrds.

Fig 7.26

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PURE MATHS IB

The areas of revolution around various lines can be obtained by integrating dS withappropriate choices of r. Below are listed some special cases

1. Suppose f ′ (x) is continuous on [a, b].(i) If the curve y = f (x) is rotated about the x−axis, the area of the surface of

revolution so generated is

S = 2π

∫ x=b

x=a|y|ds = 2π

∫ x=b

x=a|f (x)|

√1 + (f ′ (x))2dx ∵ r = |y| .

(ii) If the rotation is about the y−axis, the surface area is

S = 2π

∫ x=b

x=a|x|ds = 2π

∫ x=b

x=a|x|√

1 + (f ′ (x))2dx.

2. Suppose g′ (y) is continuous on [c, d].(i) If the curve x = g (y) is rotated about the x−axis, the area of the surface of

revolution so generated is

S = 2π

∫ y=d

y=c|y|ds = 2π

∫ y=d

y=c|y|√

1 + (g′ (y))2dy.

[(ii) If the rotation is about the y−axis, the surface area is

S = 2π

∫ y=d

y=c|x|ds = 2π

∫ y=d

y=c|g (y)|

√1 + (g′ (y))2dy.

Example 5.4.1.Find the surface area of a sphere of radius a.

Solution. Fig 7.27

Such a sphere can be generated by rotating a semicircle with equation y =√a2 − x2

−a ≤ x ≤ a about the x−axis. Thus

dy

dx= − x√

a2 − x2= −x

y.

Therefore

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PURE MATHS IB

S = 2π

∫ a

−ay

√1 +

(−xy

)2

dx

= 4π

∫ a

0

√y2 + x2dx = 4π

∫ a

0

√a2dx

= 4aπx

∣∣∣∣a0

= 4πa2 square units

Example 5.4.2 (Surface area of a parabolic dish).

Find the surface area of a parabolic reflector whose shape is obtained by rotating theparabolic arc y = x2 (0 ≤ x ≤ 1), about the y−axis.

Solution. Fig 7.28

The arc length element for the parabola y = x2 is ds =√

1 + (y′)2 =√

1 + 4x2, so the

required surface area is

s = 2π

∫ 1

0x√

1 + 4x2dx

=

[x = 1 + 4x218du = xdx

]=π

4

∫ 5

1u1/2du

=π

6u3/2

∣∣∣∣51

=π

6

(5√

5− 1)

square units

Exercise 5.4.

Find the areas of the surfaces obtained by rotating the given curve about the indicatedlines.

1. y = x3, (0 ≤ x ≤ 1), about thex−axis.

2. y = x3/2, (0 ≤ x ≤ 1), about they−axis.

3. y = sinx, (0 ≤ x ≤ π), about thex−axis.

4. y = x3

12 + 1x , (1 ≤ x ≤ 4), about the

y−axis.

65

CHAPTER 6

Polar Coordinates

6.1. Polar Coordinates and Polar Curves

Squaredbrackets used to

denote polarcoordinates

The references in a polar coordinate system are an origin (or pole) O, and a polar axis,a ray extending from O horizontally to the right. The position of a point P in the planeis then determined by the polar coordinates [r, θ], where

(i) r is the distance of P from O, and(ii) O is the angle the ray OP makes with the polar axis; counterclockwise angles

being considered positive.

Fig 8. 34

Note that polar coordinates of a point are not unique. If θ2 = θ1 + 2nπ, n ∈ Z, thenthe polar coordinates [r, θ1] and [r, θ2] represent the same point. For example, the polarcoordinates

[3,π

4

],

[3,

9π

4

],

[3,−7π

4

],

all represent the same point.If the polar and rectangular coordinate systems are considered in the same diagram

then the pole and the polar axis are chosen to coincide with the origin and the positivex−axis respectively.

Fig 8.34

The following relations immediately followx = r cos θ x2 + y2 = r2

y = r sin θ tan θ = yx

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PURE MATHS IB

Example 6.1.1.The straight line 2x− 3y = 5 has the polar equation

2c cos θ − 3r sin θ = 5 ⇒ r (2 cos θ − 3 sin θ) = 5

or r =5

2 cos θ − 3 sin θ.

Example 6.1.2.Find the Cartesian equation of the curve represented by the polar equation r = 2a cos θ;hence identify the curve.

Solution. Multiplying through by r yields r2 = 2ar cos θ which implies that x2 +y2 =2ax. This yields the equation (x− a)2 +y2 = a2. Thus the given polar equation representsa circle with center at (a, 0) and radius a.

Fig 8.35

Observe from the equation that r → 0 as θ → ±π/2.

Remark 3.square grid forCartesiansystem

The coordinate curves for the Cartesian coordinates x = constant and y = constant arevertical and horizontal lines respectively. Thus the Cartesian plane is said to be ruled byvertical and horizontal lines.

polar planeruled byconcentriccircles andradial lines

On the other hand the coordinate curves for the polar coordinates r = constant (saya) and y = constant (say β) are circles centered at the origin with radius |a| and straightlines through the origin making an angle β with polar axis respectively. Thus the polarcoordinates plane is said to be ruled by concentric circles centered at the origin and radiallines emanating from the origin.

The graph of an equation of the form r = f (θ) is called the polar graph of thefunction f .

Remark 4.

1. A polar graph r = f (θ) approaches the origin from the direction theta for whichf (theta) = 0.

2. The polar graph with equation r = f (θ − θ0) is the polar graph of of the equationr = f (θ) rotated through an angle θ0 about the origin.

Example 6.1.3.Sketch and identify the curve r = 2a cos (θ − θ0).

Solution. Ans: Circle center [a, θ0] and radius a.

Example 6.1.4.Sketch the polar curve r = a (1− cos θ) where a > 0.

Solution. Transformation to rectangular coordinate yields the equation

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PURE MATHS IB

(x2 + y2 + ax

)2= a2

(x2 + y2

)which reveals no helpful information. Therefore we make a table of values and plot somepoints.

θ 0 ±π6 ±π

4 ±π3 ±π

2 ±2π3 ±3π

4 ±5π6 π

r 0 0.13a 0.29a 0.5a a 1.5a 1.71a 1.87a 2a

Table 1

Fig 8.38

Because it is shaped like a heart, the curve is called a cardoid.

Exercise 6.1.

1. Sketch the graph of the polar equation r = 2− 2 sin θ.2. Transform the given polar equation to the rectangular coordinates, and identify

the curve represented.

(a) r = 3 sec θ.

(b) r =5

3 sin θ − 4 cos θ.

(c) r2 = csc θ.(d) r = sec θ (1 + tan θ) .

(e) r =1

1− 4 cos θ.

(f) r =2

1− 2 sin θ.

(g) r =2

1 + sin θ.

6.2. Slopes for Polar Curves

Consider a point P = [r, θ] on a polar curve other than the origin.Fig 8.50

Let Q be a point on the curve near P corresponding to the polar angle θ + h. Let S beon OQ with PS perpendicular to OQ. Then

68

PURE MATHS IB

PS = f (θ) sinh and SQ = OQ−OS = f (θ + h)− f (θ) cosh.

Let ψ be the angle the tangent line at P makes with the radial line OP . Then ψ is thelimit of the angle SQP as h→ 0. Thus

tanψ = limh→0

tan∠SQP = limh→0

PS

SQ

= limh→0

f (θ) sinh

f (θ + h)− f (θ) cosh

[0

0

]

= limh→0

f (θ) cosh

f ′ (θ + h) + f (θ) sinh(by L’Hopital’s rule)

=f (θ) · 1

f ′ (θ) + f (θ) · 0=f (θ)

f ′ (θ)=

r

drdθ.

In particular, ψ = π/2 if f ′ (θ) = 0. If ψ + θ = π, then the tangent line makes andangle 0 or π with the polar axis and so is horizontal. In this case

tan (ψ + θ) = tanπ = 0 ⇒ tanψ = − tan θ.

If ψ + θ = π2 , then the tangent line is perpendicular to the polar axis and so is vertical.

In this case

tan (ψ + θ) = tanπ

2=

1

0

⇒ tanψ + tan θ

1− tanψ tan θ=

1

0

⇒ 1− tanψ tan θ

tanψ + tan θ=

0

1= 0⇒ 1− tanψ tan θ = 0⇒ tanψ = tan θ

Remark 5.Since for parametric curves, horizontal and vertical tangents correspond to dy/dt = 0 anddx/dt = 0 respectively, it is usually easier to find the critical points of y = f (θ) sin θ forhorizontal tangents and x = f (θ) cos θ for vertical tangents.

Example 6.2.1.Find the points on the cardoid r = 1 + cos θ, where the tangent lines are vertical orhorizontal.

Solution. Multiplying the polar equation through by sin θ we obtain

r sin θ = (1 + cos θ) sin θ ⇒ y = (1 + cos θ) sin θ.

Multiplying the same equation through by cos θ we obtain

r cos θ = (1 + cos θ) cos θ ⇒ x = (1 + cos θ) cos θ.

For the horizontal tangent

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PURE MATHS IB

0 =dy

dθ= (1 + cos θ) cos θ + sin θ (− sin θ)

= · · · = 2 cos2 θ + cos θ − 1

= · · · = (2 cos θ − 1) (cos θ + 1)

⇒ cos θ =1

2or cos θ = −1

⇒θ = ±π3, θ = π

The values of r corresponding to θ = ±π3 and θ = π are 3

2 and 0 respectively. Thevalue r = 0 correspond to the origin where no tangent exists since it is a cusp. Hence thehorizontal tangents are at

[32 ,±

π3

].

For the vertical tangent

0 =dx

dθ= − (1 + cos θ) sin θ + cos θ (− sin θ)

= · · · = − sin θ (1 + 2 cos θ)

⇒ sin θ = 0 or cos θ = −1

2

⇒θ = 0, π, ±2π

3.i

The values of r corresponding to θ = 0, θ = π, and θ = ±2π3 are 2, 0 and 1

2 respectively.As already seen, when r = 0 no tangent exists. Hence the vertical tangents are at [2, 0],[12 ,±

2π3

].

Exercise 6.2.

1. Find all points on the curve r = 2 cos θ where the tangent line is horizontal,vertical or does not exist.

6.3. Area Bounded by Polar Curves

Here we determine the area A of the region R bounded by the polar graph r = f (θ)and the rays θ = α and θ = β, where we assume that β > θ and that f is continuous forα < θ < β.

Fig 8.52

The area element is thus a sector of angular width dθ so that

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PURE MATHS IB

dA =dθ

2ππr2 =

1

2r2dθ =

1

2(f (θ))2 dθ.

Thus the required area is given by

A =1

2

∫ β

α(f (θ))2dθ.

Example 6.3.1.Find the area bounded by the cardoid r = a (1 + cos θ).

Fig 8.53

Solution. By symmetry, the area is twice that of the top half;

A = 2× 1

2

∫ π

0a2 (1 + cos θ)2dθ

= a2∫ π

0

(1 + a cos θ + cos2 θ

)dθ

= a2∫ π

0

(1 + a cos θ +

1

2(1 + cos 2θ)

)dθ

= a2(θ + 2 sin θ +

1

2

(θ +

1

2sin 2θ

)) ∣∣∣∣π0

= · · · = 3

2πa2 square units.

Exercise 6.3.

Sketch and find the areas of the given polar regions R

1. R lies between the origin and the spiral r =√θ, 0 ≤ θ ≤ 2π.

2. R is bounded by the curve r2 = a2 cos 2θ.3. R is bounded by the curve r = cos 4θ.4. R lies inside the cardioid r = 1− cos θ and outside the circle r = 1.5. R lies inside the cardioid r = 1 + cos θ and outside the circle r = 3 cos θ.6. R is bounded by the smaller loop of the curve r = 1 + 2 cos θ.

7. R is enclosed by the curve r = 1 + cos θ and the radius vectors θ = 0 and θ =π

2.

71