lesson 23: the definite integral (handout)

10
Section 5.2 The Definite Integral V63.0121.006/016, Calculus I New York University April 15, 2010 Announcements I April 16: Quiz 4 on §§4.1–4.4 I April 29: Movie Day!! I April 30: Quiz 5 on §§5.1–5.4 I Monday, May 10, 12:00noon (not 10:00am as previously announced) Final Exam Announcements I April 16: Quiz 4 on §§4.1–4.4 I April 29: Movie Day!! I April 30: Quiz 5 on §§5.1–5.4 I Monday, May 10, 12:00noon (not 10:00am as previously announced) Final Exam V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 2 / 28 Objectives I Compute the definite integral using a limit of Riemann sums I Estimate the definite integral using a Riemann sum (e.g., Midpoint Rule) I Reason with the definite integral using its elementary properties. V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 3 / 28 Notes Notes Notes 1 Section 5.2 : The Definite Integral V63.0121.006/016, Calculus I April 15, 2010

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Page 1: Lesson 23: The Definite Integral (handout)

Section 5.2The Definite Integral

V63.0121.006/016, Calculus I

New York University

April 15, 2010

Announcements

I April 16: Quiz 4 on §§4.1–4.4

I April 29: Movie Day!!

I April 30: Quiz 5 on §§5.1–5.4

I Monday, May 10, 12:00noon (not 10:00am as previously announced)Final Exam

Announcements

I April 16: Quiz 4 on§§4.1–4.4

I April 29: Movie Day!!

I April 30: Quiz 5 on§§5.1–5.4

I Monday, May 10, 12:00noon(not 10:00am as previouslyannounced) Final Exam

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 2 / 28

Objectives

I Compute the definiteintegral using a limit ofRiemann sums

I Estimate the definiteintegral using a Riemannsum (e.g., Midpoint Rule)

I Reason with the definiteintegral using its elementaryproperties.

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 3 / 28

Notes

Notes

Notes

1

Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010

Page 2: Lesson 23: The Definite Integral (handout)

Outline

Recall

The definite integral as a limit

Estimating the Definite Integral

Properties of the integral

Comparison Properties of the Integral

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 4 / 28

Cavalieri’s method in general

Let f be a positive function defined on the interval [a, b]. We want to findthe area between x = a, x = b, y = 0, and y = f (x).For each positive integer n, divide up the interval into n pieces. Then

∆x =b − a

n. For each i between 1 and n, let xi be the ith step between a

and b. So

xx0 x1 xi xn−1xn. . . . . .

x0 = a

x1 = x0 + ∆x = a +b − a

n

x2 = x1 + ∆x = a + 2 · b − a

n. . .

xi = a + i · b − a

n. . .

xn = a + n · b − a

n= b

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 5 / 28

Forming Riemann sums

We have many choices of representative points to approximate the area ineach subinterval.

. . . even random points!

xIn general, choose ci to be a point in the ith interval [xi−1, xi ]. Form theRiemann sum

Sn = f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x =n∑

i=1

f (ci )∆x

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 6 / 28

Notes

Notes

Notes

2

Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010

Page 3: Lesson 23: The Definite Integral (handout)

Theorem of the (previous) Day

Theorem

If f is a continuous function on [a, b]or has finitely many jumpdiscontinuities, then

limn→∞

Sn = limn→∞

{n∑

i=1

f (ci )∆x

}

exists and is the same value nomatter what choice of ci we make. x

M15 = 7.49968

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 7 / 28

Outline

Recall

The definite integral as a limit

Estimating the Definite Integral

Properties of the integral

Comparison Properties of the Integral

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 8 / 28

The definite integral as a limit

Definition

If f is a function defined on [a, b], the definite integral of f from a to bis the number ∫ b

af (x) dx = lim

∆x→0

n∑i=1

f (ci ) ∆x

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 9 / 28

Notes

Notes

Notes

3

Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010

Page 4: Lesson 23: The Definite Integral (handout)

Notation/Terminology

∫ b

af (x) dx = lim

∆x→0

n∑i=1

f (ci ) ∆x

I

∫— integral sign (swoopy S)

I f (x) — integrand

I a and b — limits of integration (a is the lower limit and b theupper limit)

I dx — ??? (a parenthesis? an infinitesimal? a variable?)

I The process of computing an integral is called integration orquadrature

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 10 / 28

The limit can be simplified

Theorem

If f is continuous on [a, b] or if f has only finitely many jumpdiscontinuities, then f is integrable on [a, b]; that is, the definite integral∫ b

af (x) dx exists.

Theorem

If f is integrable on [a, b] then∫ b

af (x) dx = lim

n→∞

n∑i=1

f (xi )∆x ,

where

∆x =b − a

nand xi = a + i ∆x

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 11 / 28

Example: Integral of x

Example

Find

∫ 3

0x dx

Solution

For any n we have ∆x =3

nand xi =

3i

n. So

Rn =n∑

i=1

f (xi ) ∆x =n∑

i=1

(3i

n

)(3

n

)=

9

n2

n∑i=1

i

=9

n2· n(n + 1)

2−→ 9

2

So

∫ 3

0x dx =

9

2= 4.5

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 12 / 28

Notes

Notes

Notes

4

Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010

Page 5: Lesson 23: The Definite Integral (handout)

Example: Integral of x2

Example

Find

∫ 3

0x2 dx

Solution

For any n we have ∆x =3

nand xi =

3i

n. So

Rn =n∑

i=1

f (xi ) ∆x =n∑

i=1

(3i

n

)2(3

n

)=

27

n3

n∑i=1

i2

=27

n3· n(n + 1)(2n + 1)

6−→ 27

3= 9

So

∫ 3

0x2 dx = 9

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 13 / 28

Example: Integral of x3

Example

Find

∫ 3

0x3 dx

Solution

For any n we have ∆x =3

nand xi =

3i

n. So

Rn =n∑

i=1

f (xi ) ∆x =n∑

i=1

(3i

n

)3(3

n

)=

81

n4

n∑i=1

i3

=81

n4· n2(n + 1)2

4−→ 81

4

So

∫ 3

0x3 dx =

81

4= 20.25

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 14 / 28

Outline

Recall

The definite integral as a limit

Estimating the Definite Integral

Properties of the integral

Comparison Properties of the Integral

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 15 / 28

Notes

Notes

Notes

5

Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010

Page 6: Lesson 23: The Definite Integral (handout)

Estimating the Definite Integral

Example

Estimate

∫ 1

0

4

1 + x2dx using M4.

Solution

We have x0 = 0, x1 =1

4, x2 =

1

2, x3 =

3

4, x4 = 1.

So c1 =1

8, c2 =

3

8, c3 =

5

8, c4 =

7

8.

M4 =1

4

(4

1 + (1/8)2+

4

1 + (3/8)2+

4

1 + (5/8)2+

4

1 + (7/8)2

)=

1

4

(4

65/64+

4

73/64+

4

89/64+

4

113/64

)=

64

65+

64

73+

64

89+

64

113≈ 3.1468

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 16 / 28

Estimating the Definite Integral (Continued)

Example

Estimate

∫ 1

0

4

1 + x2dx using L4 and R4

Answer

L4 =1

4

(4

1 + (0)2+

4

1 + (1/4)2+

4

1 + (1/2)2+

4

1 + (3/4)2

)= 1 +

16

17+

4

5+

16

25≈ 3.38118

R4 =1

4

(4

1 + (1/4)2+

4

1 + (1/2)2+

4

1 + (3/4)2+

4

1 + (1)2

)=

16

17+

4

5+

16

25+

1

2≈ 2.88118

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 17 / 28

Outline

Recall

The definite integral as a limit

Estimating the Definite Integral

Properties of the integral

Comparison Properties of the Integral

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 18 / 28

Notes

Notes

Notes

6

Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010

Page 7: Lesson 23: The Definite Integral (handout)

Properties of the integral

Theorem (Additive Properties of the Integral)

Let f and g be integrable functions on [a, b] and c a constant. Then

1.

∫ b

ac dx = c(b − a)

2.

∫ b

a[f (x) + g(x)] dx =

∫ b

af (x) dx +

∫ b

ag(x) dx.

3.

∫ b

acf (x) dx = c

∫ b

af (x) dx.

4.

∫ b

a[f (x)− g(x)] dx =

∫ b

af (x) dx −

∫ b

ag(x) dx.

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 19 / 28

Proofs

Proofs.

I When integrating a constant function c , each Riemann sum equalsc(b − a).

I A Riemann sum for f + g equals a Riemann sum for f plus aRiemann sum for g . Using the sum rule for limits, the integral of asum is the sum of the integrals.

I Ditto for constant multiples

I Ditto for differences

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 20 / 28

Example

Find

∫ 3

0

(x3 − 4.5x2 + 5.5x + 1

)dx

Solution

∫ 3

0(x3−4.5x2 + 5.5x + 1) dx

=

∫ 3

0x3 dx − 4.5

∫ 3

0x2 dx + 5.5

∫ 3

0x dx +

∫ 3

01 dx

= 20.25− 4.5 · 9 + 5.5 · 4.5 + 3 · 1 = 7.5

(This is the function we were estimating the integral of before)

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 21 / 28

Notes

Notes

Notes

7

Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010

Page 8: Lesson 23: The Definite Integral (handout)

Theorem of the (previous) Day

Theorem

If f is a continuous function on [a, b]or has finitely many jumpdiscontinuities, then

limn→∞

Sn = limn→∞

{n∑

i=1

f (ci )∆x

}

exists and is the same value nomatter what choice of ci we make. x

M15 = 7.49968

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 22 / 28

More Properties of the Integral

Conventions: ∫ a

bf (x) dx = −

∫ b

af (x) dx

∫ a

af (x) dx = 0

This allows us to have

5.

∫ c

af (x) dx =

∫ b

af (x) dx +

∫ c

bf (x) dx for all a, b, and c.

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 23 / 28

Example

Suppose f and g are functions with

I

∫ 4

0f (x) dx = 4

I

∫ 5

0f (x) dx = 7

I

∫ 5

0g(x) dx = 3.

Find

(a)

∫ 5

0[2f (x)− g(x)] dx

(b)

∫ 5

4f (x) dx .

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 24 / 28

Notes

Notes

Notes

8

Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010

Page 9: Lesson 23: The Definite Integral (handout)

Solution

We have

(a) ∫ 5

0[2f (x)− g(x)] dx = 2

∫ 5

0f (x) dx −

∫ 5

0g(x) dx

= 2 · 7− 3 = 11

(b) ∫ 5

4f (x) dx =

∫ 5

0f (x) dx −

∫ 4

0f (x) dx

= 7− 4 = 3

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 25 / 28

Outline

Recall

The definite integral as a limit

Estimating the Definite Integral

Properties of the integral

Comparison Properties of the Integral

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 26 / 28

Comparison Properties of the Integral

Theorem

Let f and g be integrable functions on [a, b].

6. If f (x) ≥ 0 for all x in [a, b], then∫ b

af (x) dx ≥ 0

7. If f (x) ≥ g(x) for all x in [a, b], then∫ b

af (x) dx ≥

∫ b

ag(x) dx

8. If m ≤ f (x) ≤ M for all x in [a, b], then

m(b − a) ≤∫ b

af (x) dx ≤ M(b − a)

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 27 / 28

Notes

Notes

Notes

9

Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010

Page 10: Lesson 23: The Definite Integral (handout)

Example

Estimate

∫ 2

1

1

xdx using the comparison properties.

Solution

Since1

2≤ x ≤ 1

1

for all x in [1, 2], we have

1

2· 1 ≤

∫ 2

1

1

xdx ≤ 1 · 1

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 28 / 28

Notes

Notes

Notes

10

Section 5.2 : The Definite IntegralV63.0121.006/016, Calculus I April 15, 2010