© boardworks ltd 2006 1 of 57 © boardworks ltd 2006 1 of 57 a-level maths: core 4 for edexcel c4.5...
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A-Level Maths: Core 4for Edexcel
C4.5 Integration 1
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For more detailed instructions, see the Getting Started presentation.
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Integrals of standard functions
Reversing the chain rule
Integration by substitution
Integration by parts
Volumes of revolution
Examination-style question
Integrals of standard functions
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Review of integration
So far, we have only looked at functions that can be integrated using:
1
= + for all 11
nn kx
kx dx c nn
For example:
Integrate with respect to x.42
110 + + 6 3x x
x
42
110 + + 6 3 =x x dx
x
124 210 + + 6 3x dx x dx x dx dx
5 31= 2 + 4 3 +x x x c
x
510=
5
x 1
+1
x
32
32
6+
x3x + c
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The integral of
Adding 1 to the power and then dividing would lead to the meaningless expression,
The only function of the form xn that cannot be integrated by this method is x–1 = .1
x
0
0
x
This does not mean that cannot be integrated.1x
Remember that 1
if = ln then =dy
y xdx x
Therefore 1
= ln + (where > 0)dx x c xx
1x
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The integral of
We can only find the log of a positive number and so this is only true for x > 0.
We can get around this by taking x to be negative.
However, does exist for x < 0 (but not x = 0). So how do we integrate it for all possible values of x?
1x
If x < 0 then –x > 0 so:
1 1 = = ln( )+ (where < 0)dx dx x c x
x x
1 = ln + dx x c
x
We can combine the integrals of for both x > 0 and x < 0 by using the modulus sign to give:
1x
1x
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The integral of
Find 2
3dx
x
This is just the integral of multiplied by a constant. 1x
2=
3dx
x2 1
=3
dxx
2ln +
3x c
Find 22
54x x dx
x
22
54 =x x dx
x 3 20
4x dxx
4= 20ln +x x c
1x
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Definite integrals involving
It is particularly important to remember the modulus sign when evaluating definite integrals of functions involving . 1
x
Find the area under the curve y = – between x = –3, x = –1 and the x-axis, writing your answer in the form ln a.
1x
0 x
y
–1–3
1y
x
The area is given by .1
3
1
x
1 1
33
1= ln x
x
= ln 1 ln 3
= ln1+ ln3 Remember that ln 1 = 0
1x
= ln3 units squared
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Definite integrals involving
We should note that definite integrals of the form can only1b
a xbe evaluated if x = 0 does not lie in the interval [a, b].
1x
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1
= + ( 1)1
nn x
x dx c nn
Integrals of standard functions
By reversing the process of differentiation we can derive the integrals of some standard functions.
1 = ln + dx x c
x
= + x xe dx e c
cos = sin + x dx x c
sin = cos +x dx x c
2sec = tan + x dx x c
These integrals should be memorized.
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Integrals of standard functions
Also, if any function is multiplied by a constant k then its integral will also be multiplied by the constant k.
Find 4sin + 7 xx e dx 4sin + 7 = 4sin + 7x xx e dx x dx e dx
= 4 sin + 7 xx dx e dx = 4( cos )+ 7 xx e
= 7 4cosxe x
In practice most of these steps can be left out.
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Integrals of standard functions
Reversing the chain rule
Integration by substitution
Integration by parts
Volumes of revolution
Examination-style question
Reversing the chain rule
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Reversing the chain rule
A very helpful technique is to recognize that a function that we are trying to integrate is of a form given by the differentiation of a composite function. This is sometimes called integration by recognition.
Let f +1= ( ( ))ny x
By the chain rule: f f= ( +1)( ( )) '( )ndyn x x
dx
So f f f +1( +1)( ( )) '( ) = ( ( )) +n nn x x dx x cIt follows that for n ≠ 1
f f f +11( ( )) '( ) = ( ( )) +
( +1)n nx x dx x c
n
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Reversing the chain rule
Suppose we want to integrate (2x + 7)5 with respect to x.
If the integral is multiplied by a constant k:
Consider the derivative of y = (2x + 7)6.
Using the chain rule: 5= 6(2 + 7) ×2dy
xdx
= 12(2x + 7)5
So5 612(2 + 7) = (2 + 7) +x dx x c5 61
(2 + 7) = (2 + 7) +12
x dx x c
f f f +1( ( )) '( ) = ( ( )) +( +1)
n nkk x x dx x c
n
Don’t try to learn this formula, just try to recognize that the function you are integrating is of the form k(f(x))n f ’(x) and compare it to the derivative of (f(x))n + 1.
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Reversing the chain rule
In general, you can integrate any linear function raised to a power using the formula:
+11( ) = ( ) +
( +1)
n nax b dx ax b ca n
With practice, integrals of this type can be written down directly. For example:
8( 5) =x dx 91( 5) +
9x c
2(4 + 7) =x dx 31(4 + 7) +
12x c
95(3 2 ) =x dx 105(3 2 ) + =
20x c
101
(3 2 ) +4
x c
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Reversing the chain rule
Integrate y = x(3x2 + 4)3 with respect to x.
Notice that the derivative of 3x2 + 4 is 6x.
Using the chain rule: 2 3= 4(3 + 4) ×6dy
x xdx
= 24x(3x2 + 4)3
So2 3 2 424 (3 + 4) = (3 + 4) +x x dx x c2 3 2 41
(3 + 4) = (3 + 4) +24
x x dx x c
Let’s look at some more integrals of functions of the form k(f(x))n f ’(x).
Now consider the derivative of y = (3x2 + 4)4.
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Reversing the chain rule
Now consider the derivative of y = (2x3 – 9)3.
Using the chain rule: 3 2 2= 3(2 9) ×6dy
x xdx
= 18x2(2x3 – 9)2
So2 3 2 3 318 (2 9) = (2 9) +x x dx x c 2 3 2 3 31(2 9) = (2 9) +
18x x dx x c
Find . 2 3 27 (2 9)x x dx
Notice that the derivative of 2x3 – 9 is 6x2.
2 3 2 3 377 (2 9) = (2 9) +
18x x dx x c
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Reversing the chain rule
So1 12 22 3 33
( 1) = ( 1) +2
x x dx x c 1 12 22 3 32
( 1) = ( 1) +3
x x dx x c
Find .
2
3 1
xdx
x
Start by writing as 2
3 1
x
x
122 3( 1) .x x
Now consider the derivative of y =123( 1) .x
x2 is the derivative of (x3 – 1).
plus 1 is12 1
2 .
Using the chain rule:123 21
= ( 1) ×3 =2
dyx x
dx
122 33
( 1)2
x x
2 3
3
2 1= +
31
x xdx c
x
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Reversing the chain rule for exponential functions
When we applied the chain rule to functions of the form ef(x) we obtained the following generalization:
We can reverse this to integrate functions of the form k f ’(x)ef(x). For example:
f ff( ) ( )If = then = '( )x xdyy e x e
dx
5 35 =xe dx 5 3 +xe c
A numerical adjustment is usually necessary.
5 3 =xe dx 5 31+
5xe c
5 32 =xe dx 5 32+
5xe c
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Reversing the chain rule for exponential functions
In general, 1= +ax b ax be dx e c
a
Find . 3
6x
dxe
36= +
3xe c
33
6= 6 x
xdx e dx
e
3= 2 +xe c
3
2= +
xc
e
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Reversing the chain rule for exponential functions
With practice, this method can be extended to cases where the exponent is not linear. For example:
Find .
22xxe dx
Notice that the derivative of 2x2 is 4x and so the function we are integrating is of the form k f ’(x)ef(x).
221= +
4xe c
22 =xxe dx221
44
xxe dx
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Reversing the chain rule for logarithmic functions
When we applied the chain rule to functions of the form ln f(x) we obtained the following generalization:
ff
f
'( )If = ln ( ) then =
( )
dy xy x
dx x
We can reverse this to integrate functions of the form For example:
f
f
'( )
( )
k x
x
1=
5 + 4dx
x1
= ln 5 + 4 +5
x c
In general, 1 1= ln + +
+dx ax b c
ax b a
1 5
5 5 + 4dx
xRemove a factor of to write the function in the form .
15
ff'( )( )
xx
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Reversing the chain rule for logarithmic functions
Find . 5
2 6dx
x5
=2 6
dxx
5 6
6 2 6dx
x
5
= ln 2 6 +6
x c
Evaluate , writing your answer in the form a ln b. 2
1
4
2 7dx
x
First of all, note that the graph of y = has a discontinuity
when 2x – 7 = 0, that is when x = 3.5.
4
2 7x
This is outside the interval [–1, 2] and so the integral is valid.
This is now of the form .f
f'( )( )
xx
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Reversing the chain rule for logarithmic functions
2 2
1 1
4 2= 2
2 7 2 7dx dx
x x
2
1= 2 ln 2 7x
= 2(ln 4 7 ln 2 7 )
= 2(ln3 ln9)
This can be written in the required form by using the rule that ln a – ln b = ln .
a
b2
1
4 3= 2ln
2 7 9dx
x
This is now of the form .f
f'( )( )
xx
1= 2ln
3
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Reversing the chain rule for logarithmic functions
Find . 2
37 + 2
xdx
x2
3=
7 + 2
xdx
x2
3
1 6
6 7 + 2
xdx
x
31= ln 7 + 2 +
6x c
This is now of the form .f
f'( )( )
xx
Find . 2
2
4
+1
x
x
edx
e2
2
4=
+1
x
x
edx
e2
2
22
+1
x
x
edx
e
2= 2ln +1 +xe c
This is now of the form .f
f'( )( )
xx
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The integral of tan x
We can find the integral of tan x by writing it as and
recognizing that this fraction is of the form .
sin
cos
x
xf
f
'( )
( )
x
x
sintan =
cos
xx dx dx
x = ln cos +x c
It is ‘tidier’ to rewrite this without a minus sign at the front, using the fact that –ln a = ln a–1:
1ln cos + = ln cos +x c x c
= ln sec +x c
1= ln +
cosc
x
tan = ln sec +x dx x c
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Reversing the chain rule for trigonometric functions
When we applied the chain rule to functions of the form sin f(x) and cos f(x) we obtained the following generalizations:
We can reverse these to integrate functions of the form f ’(x) cos f(x) and f ’(x) sin f(x). For example:
f f fIf = sin ( ) then = '( )cos ( )dy
y x x xdx
f f fIf = cos ( ) then = '( )sin ( )dy
y x x xdx
3cos3 =x dx sin3 +x c
22 sin =x x dx 2cos +x c
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Reversing the chain rule for trigonometric functions
As with other examples a numerical adjustment is often necessary.
In general, when dealing with the cos and sin of linear functions:
sin(5 2) =x dx1
5sin(5 2)5
x dx This is now of the form –f ’(x) sin f(x).
1= cos(5 2)
5x dx
1cos( + ) = sin( + )+ax b dx ax b c
a
1sin( + ) = cos( + )+ax b dx ax b c
a
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Integrals of standard functions
Reversing the chain rule
Integration by substitution
Integration by parts
Volumes of revolution
Examination-style question
Integration by substitution
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Integration by substitution
With practice, the technique of integration by recognition can save a lot of time.
However, when it is too difficult to use integration by recognition we can use a more formal method of reversing the chain rule called integration by substitution.
To see how this method works consider the integral 3
(5 + 2) .x dxLet u = 5x + 2 so that
3(5 + 2) =x dx 3u dx
The problem now is that we can’t integrate a function in u with respect to x. We therefore need to write dx in terms of du.
= 5 + 2 = 5du
u xdx
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Integration by substitution
Now change the variable back to x:
When we used the chain rule for differentiation we saw that we can treat informally as a fraction, so:
du
dx
= 5 = 5du
du dxdx
1
=5
dx du
So if u = 5x + 2 and dx : 1
=5
du
3( )5 + 2 =x dx 3 1
5u du
41= +
20u c
3 41(5 + 2) = (5 + 2) +
20x dx x c
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Reversing the chain rule for logarithmic functions
Use a suitable substitution to find . 2 5(2 5)x x dx
Let u = 2x2 – 5 = 4du
xdx
1=
4
dx
du x1
=4
dx dux
Substituting u and dx into the original problem gives:
Notice that the x’s cancel out.
2 52 5( ) =x x dx 5 1
4ux du
x51
=4
u du61
= +24
u c
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Integration by substitution
This integral could also have been found directly by recognition.
Now we need to change the variable back to x :
2 5 2 61(2 5) = (2 5) +
24x x dx x c
However, there are functions that can be integrated by use of a suitable substitution but not by recognition. For instance:
Use the substitution u = 1 – 2x to find . 4(1 2 )x x dx
If u = 1 – 2x then = 2du
dx
1=
2
dx
du 1
=2
dx du
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Integration by substitution
Substituting these into the original problem gives:
We also have to substitute the x so that the whole integrand is in terms of u.
4( )1 2 =xx dx 4 1(
1(1 )
2)
2duuu
41= (1 )
4u u du
Also if u = 1 – 2x then1
= (1 )2
x u
4 51= ( )
4u u du
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Integration by substitution
5 64 51 1
( ) = +4 4 5 6
u uu u du c
Changing the variable back to x gives:
4(1 2 ) =x x dx 51(1 2 ) (6 5(1 2 ))+
120x x c
51= (1 2 ) (10 +1)+
120x x c
5 61 6 5= +
4 30
u uc
51= (6 5 )+
120u u c
51= (2 1) (10 +1)+
120x x c
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Definite integration by substitution
When a definite integral is found by substitution it is easiest to rewrite the limits of integration in terms of the substituted variable.
12
1= (8 ) ( 1)
2
dux
dx If u = then
12(8 )x
Using the chain rule for differentiation.
12
1=
2(8 )x
1
=2u
So = 2dx
udu
= 2dx u du
Use the substitution u = to find the area under the curve8 x
y = between x = 4 and x = 7. 2
8
x
x
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Definite integration by substitution
Rewrite the limits in terms of u:
Now we need to find x in terms of u.
u2 = 8 – x
x = 8 – u2
= 8 = 24u when x = 3,
= 8 =17u when x = 1,
7
4
2=
8
x
xdx
2 2
1
2( 2
))
(8 u
uu du
2 2
1= 4(8 )u du
2 2
1= 4 ( 8)u du
The area is given by . Rewrite this in terms of u:7
4
2
8
xdx
x
12(8 )xIf u = then
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Definite integration by substitution
232 2
11
4 ( 8) = 4 83
uu du u
8 1= 4 16 + 8
3 3
7= 4 8
3
23= 22
Therefore, the required area is units squared.2322
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Co
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Integrals of standard functions
Reversing the chain rule
Integration by substitution
Integration by parts
Volumes of revolution
Examination-style question
Integration by parts
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Integration by parts
Suppose we wish to integrate the product of two functions, such as x sin x, where one of the functions is not related to the derivative of the other.
An expression such as this can be integrated using the method of integration by parts.When we differentiate the product of two functions u and v we use the product rule:
( ) = +d dv du
uv u vdx dx dx
Integrating throughout with respect to x gives:
where u and v are functions of x.
= +dv du
uv u dx v dxdx dx
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Integration by parts
This can be rearranged to give:
It is important to choose u and so that
=dv du
u dx uv v dxdx dx
To integrate a product using this formula we let one part equal
u and the other equal . dv
dx
We find by differentiating the part we called u. du
dx
We find v by integrating the part we called .dv
dx
duv dx
dxdv
u dxdxis easier to integrate than .
dv
dx
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Integration by parts
So, to integrate x sin x with respect to x:
Let and=u x = sindv
xdx
So =1du
dxWe don’t need the “+ c” here.Now, using the formula : =
dv duu dx uv v dx
dx dx
sin = cos cosx x dx x x x dx = cos sin +x x x c
= sin cos +x x x c
differentiate integrate
= cosv xand
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Integration by parts
So and=1du
dx
2
=2
xev
Now, apply the formula : =dv du
u dx uv v dxdx dx
2 2
2 =2 2
x xx e e
xe dx x dx
Find .2xxe dx
2 2
= +2 4
x xxe ec
2
= (2 1)+4
xex c
2= xdve
dxLet and =u x
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Integration by parts
let and = lnu x = 8dv
xdx
so and1
=du
dx x2= 4v x
Now, using the formula, : =dv du
u dx uv v dxdx dx
2 2 1
8 ln = 4 ln 4 ×x x dx x x x dxx
Find .8 lnx x dxWe don’t know the integral of ln x so:
2= 4 ln 4x x x dx 2 2= 4 ln 2 +x x x c2= 2 (2ln 1)+x x c
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The integral of ln x
We can also use integration by parts to find the integral of ln x.
We do this by writing ln x as (1 × ln x).
Let and = lnu x =1dv
dx
So 1
=du
dx x
Now we can integrate by parts:1
1×ln = lnx dx x x x dxx
= ln 1x x dx = ln +x x x c
ln = ln +x dx x x x c
=v xand
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Integration by parts
To evaluate a definite integral using integration by parts we use:
Let and = 2u x = cosdv
xdx
So = 2du
dx
=b bb
aa a
dv duu dx uv v dx
dx dx
Evaluate .2
02 cosx x dx
2 22
00 02 cos = 2 sin 2sinx x dx x x x dx
2
02= ( sin 0) 2cos x
2= ( 2cos + 2cos0) sin π/2 = 1
cos π/2 = 0cos 0 = 1 = 2
= sinv xand
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Co
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Integrals of standard functions
Reversing the chain rule
Integration by substitution
Integration by parts
Volumes of revolution
Examination-style question
Volumes of revolution
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Volumes of revolution
Consider the area bounded by the curve y = f(x), the x-axis and x = a and x = b.
If this area is rotated 360° about the x-axis a three-dimensional shape called a solid of revolution is formed.
The volume of this solid is called its volume of revolution.
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Volumes of revolution
We can calculate the volume of revolution by dividing the volume of revolution into thin slices of width δx.
The volume of each slice is approximately cylindrical, of radius y and height δx, and is therefore approximately equal to
πy2δx
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Volumes of revolution
The total volume of the solid is given by the sum of the volume of the slices. =
2
=
x b
x a
V y x
The smaller δx is, the closer this approximate area is to the actual area.
We can find the actual area by considering the limit of this sum as δx tends to 0. =
2
0=
= limx b
xx a
V y x
This limit is represented by the following integral:
2=b
aV y dx
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Volumes of revolution
So in general, the volume of revolution V of the solid generated by rotating the curve y = f(x) between x = a and x = b about the x-axis is:
Similarly, the volume of revolution V of the solid generated by rotating the curve x = f(y) between y = a and y = b about the y-axis is:
Volumes of revolution are usually given as multiples of π.
2=b
aV y dx
2=b
aV x dy
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Volumes of revolution
Find the volume of the solid formed by rotating the area between the curve y = x(2 – x), the x-axis, x = 0, and x = 2 360° about the x-axis.
2 2
0=V y dx
2 2 2
0= (2 )x x dx
2 2 3 4
0= (4 4 )x x x dx
23 4 5
0
4 4=
3 4 5
x x x
32 32= 16 +
3 5
16cubic u s
15= nit
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Volumes of revolution
2 2
1=V x dy
22
1
1= dy
y
2 2
1= y dy
2
1
1=
x
1= +1
2
cubic u= nits
2
Find the volume of the solid formed by rotating the area between the curve y = , the y-axis, y = 1, and y = 2 360° about the y-axis.
1x
Rearranging y = gives x = .
1x
1y
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Integrals of standard functions
Reversing the chain rule
Integration by substitution
Integration by parts
Volumes of revolution
Examination-style question
Examination-style question
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Examination-style question
a) Show that cos2x = (1 + cos 2x)
b) The diagram below shows the finite region R which is bounded by the curve y the x-axis, the y-axis and the line x = .
Using the substitution x = sin u, or otherwise, find the area of the region R to 3 significant figures.
c) Find the volume of the solid formed when R is rotated through 360° about the x-axis to 3 significant figures.
12
12
2= 1 x
12
x
y
0
R
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Examination-style question
a) RHS = (1 + cos 2x)12
= (1 + (2cos2x – 1))12
= (2cos2x)12
= cos2x = LHS
Using the double angle identity cos 2A ≡ 2cos2A – 1.
b) The area of R is given by12 2
01 x dx
Using the substitution x = sin u we have:
Find u when x is 0 and ½:
= cos = cosdx
u dx u dudu
when x = 0, u = 0 and when x = , u =12 6
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Examination-style question
Now make the substitution:12 2
01 =x dx
6 2
0= cos u du
6
0
12= (1+ cos2 )u du
Using the identity cos2A = (1 + cos 2A).1
2
6
0
1 12 2= + sin2u u
6 2
01 sin cosu u du
3= +
12 8
Therefore, the area of region R is 0.478 units squared (to 3 s.f.).
Using the identity cos2A = 1 – sin2A.
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Examination-style question
181
=2 3
11=
24
Therefore, the volume formed when R is rotated about the x-axis is 1.44 units cubed (to 3 s.f.).
c) The required volume of revolution is given by 12 2
0=y dx
123
0
=3
xx
12 2
0(1 )x dx