Lecture 10. Integration (continued)
 Integration by substitution – definite integrals
 Integration by parts
 Integration of trigonometric functions
 Applications of integrations
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Definite integral by substitution
Evaluate
ò
t
4
3t +1
2
1
dt
Solution
du
du
= 6t, which gives tdt =
dt
6
We have to substitute values for the limits t = 1 and t = 4, because we are
integrating with respect to the new variable u = 3t2 + 1
Let u = 3t 2 +1. Hence,
When t = 1, u = 3 x 12 + 1 = 4
When t = 4, u = 3 x 42 +1 = 49
ò
4
1
t
3t +1
2
dt =
ò
u=49
u=4
1 du 1
=
u 6 6
ò
1/2 ù
é
1
u
1 1/2 1/2 5
u-1/2 du = ê
=
(49 - 4 ) =
ú
6 ë1/ 2 û4 3
3
49
49
4
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Example
Evaluate
ò
p /2
0
cos x sin x dx
Solution
Let u = sin x. Hence,
du
= cos x Þ cos xdx = du
dx
We have to change the limits of integration:
ò
p /2
0
cos x sin x dx =
ò
u=1
u=0
u du = ò
x = 0, u = sin(0) = 0;
x = π/2, u = sin(π/2) = 1
é u3/2 ù 2 3/2 3/2 2
u du = ê
ú = (1 - 0 ) =
3
ë 3 / 2 û0 3
1
1
0
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Integration by parts
The product rule for differentiation of two functions u(x) and v(x) is given by
d
(uv) = u¢v + uv¢
dx
Integrating both sides gives
Rearranging yields
uv = ò u¢vdx + ò uv¢ dx
ò uv¢ dx = uv - ò u¢vdx
it is called the integration by parts formula
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ò uv¢ dx = uv - ò u¢vdx
Example
Find
x
xe
ò dx
Solution
Let u = x, v′ = ex. Then
v = ò ex dx = ex; u¢ = 1
Substitute everything into the formula
x
x
x
x
x
x
xe
dx
=
xe
1×e
dx
=
xe
e
+
C
=
e
(x -1) + C
ò
ò
Why have we chosen u = x, v′ = ex and not other way round?
If we had selected v′ = x, u = ex , then
1 2
v = x ; u¢ = ex
2
1 2 x
x2 x
ò xe dx = 2 x e - ò 2 e dx even more complicated than the original one
x
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ò uv¢ dx = uv - ò u¢vdx
Example
Find
ò xln xdx
Solution
Let
u = ln x
1
u¢ =
x
v¢ = x
x2
v = ò xdx =
2
x2
1 x2
x2
1
x2
1 x2
òxln xdx = ln x 2 - òx 2 dx = 2 ln x - 2 òxdx = 2 ln x - 2 ò 2
x2 ò
1ò
= òln x - ò + C
2ò
2ò
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ò uv¢ dx = uv - ò u¢vdx
Example
Find
ò t cost dt
Solution
Let
u= t
v¢ = cost
u¢ = 1
v = ò cost dt = sin t
ò t cost dt = t sint - ò sint dt = t sint + cost + C
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Integration by parts formula for definite integrals
ò
b
a
uv¢ dx = [uv]a - ò u¢vdx
b
b
a
Example
1
1
1
1
1
1
1



2
t

2
t

2
0
2t
te dt   te

(

e
)
dt


(1
e

0
e
)

e
dt
 2

0
2
2 0

0 0 2

1
2t


1
1 2 1  1 2t 
1 2 1 2 0
3 2 1 1
  e   e    e  (e  e )   e   1  3e2
2
2 2
2
4
4
4 4
0

u= t
v¢ = e-2t
u¢ = 1
1
v = ò e-2tCE1MAT
dt = - e-2t
2

8
Example (mechanics)
x of a particle is given by x = te-t
Find the velocity, v = x for the initial condition t = 0, v = 0.
The acceleration,
Solution
v = x = ò te-t dt


 tet  (et )dt  tet  et dt  tet  et  C
u= t
v¢ = e-t
u¢ = 1
v = ò e-t dt = -e-t
Substituting the given conditions t = 0, v = 0, we have
Hence,
0  0e0  e0  C  C  1
v = -te-t - e-t +1 = 1- e-t (1+ t)
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Displacement, velocity and acceleration relation
The displacement, s, velocity, v, and acceleration a,
are related by
ò vdt
v = ò adt
s=
Moreover, if we are given initial conditions, we can find
the value of the integration constant C.
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Example
The pressure, P at a distance x of a fluid flowing around a sphere of radius r is
given by
x
ò1+ r 3 / x3 ò
P = ò kò
-ò ò
x
4
dx
ò
ò
where k is a constant. Evaluate P.
Solution
ò1+ r 3 / x3 ò
P = ò kò
òdx =
4
-ò
ò x
ò
x
x
ò 1 r 3 / x3 ò
x 1
x 1
3
k
+
dx
=
k
dx
+
kr
ò-ò òòx4 x4 òò
ò-ò x4
ò-ò x7 dx
x
x
òx ò
k ò1 ò
kr 3 ò 1 ò
k
kr 3
3 òx ò
= k ò ò + kr ò ò = - ò 3 ò =- 3 - 6
6ò
ò
3 òx ò-ò 6 òx ò-ò
3x 6x
ò-3 ò-ò
ò-6 ò-ò
-3
-6
x
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More examples
ò
ln p
0
ex sin(ex )dx
du
= ex Þ exdx = du
dx
x = 0, u = e0 = 1
Let u = ex Þ
x = ln p , u = eln p = p - new limits
=
ò
p
1
sin(u) du =
ò
p
1
p
é
ù
sin(u)du = -ëcosuû = -(cos p - cos1) =1.54
1
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Example of using integration by parts twice
Find
ò
1
-1

1
2 x
x e dx
2 x 1
x e dx   x e  
1
1
2 x
u= x
v¢ = e
u¢ = 1
v = ex

1

1
1
v¢ = ex
u¢ = 2x
v = ex
2 xe dx  (e  e )  2
x
1
x 1
xe dx   xe  
1
1
x
1
u = x2
x

1
1

1
1
xe x dx
e x dx
1
 (e  (1)e )  e x   (e1  e 1 )  (e1  e 1 )  2e 1
1
1
ò
1
-1
1
x2 ex dx = e- e-1 - 2 × 2e-1 = e- 5e-1
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Integration of trigonometric functions
1
sin x  (1  cos 2 x)
2
2
 cos xdx  sin x  C

1
sin 2 xdx   cos 2 x  C
2
 sin
2
xdx


1
1
1
1


(1  cos 2 x)dx 
dx  cos 2 xdx   x  sin 2 x  C 
2
2
2
2

1
1
 x  sin 2 x  K , where K  C / 2
2
4

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
14
Example
Find
sin2x = 2sin xcos x
cos2x
ò cos2 xsin2 x dx
Solution
cos2x
4cos2x
dx
=
ò cos2 xsin2 x ò sin2 2x dx
du
du
Let u = sin 2x Þ
= 2cos2x Þ cos2xdx =
dx
2
ò
4cos2x
dx =
2
sin 2x
ò
4 du
du
2
-2
-1
=
2
=
2
u
du
=
-2u
+
C
=
+C
ò
ò
2
2
u 2
u
sin 2x
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1
sin x cos x  sin 2 x
2
Example
Find the area under the curve sin(x)cos(x) for x between 0 and 2π

2
0
sin x cos xdx


2
0
1
1
1
2
sin 2 xdx    cos 2 x 0    cos 4  cos0 
2
4
4
1
  (1  1)  0
4
The correct value of the definite
integral is 0,
But the correct value of the total
area enclosed is
+
+
4
-
-

 /2
0
sin x cos xdx  4 
 (cos   cos0)  2
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cos 2 x0 /2
4
16
Engineering applications
Area =
b
ò ydx, where y = f (x)
a
We can compare exact and
approximate evaluations of the
integral by determining the
percentage error:
æ exact - approximate ö
% error = ç
÷ ´100
è
ø
exact
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Example
Area =
1
ò
0
é1 ù 1
x2 dx = ê x3 ú =
ë 3 û0 3
Number of intervals
1
òx
2
dx (trapezium rule)
0
Percentage error
1
4
8
16
32
64
0.34375
0.33594
0.33398
0.33349
0.33337
-3.125
-0.782
-0.194
-0.047
-0.011
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Simpson’s rule
To calculate area under the curve with the Simpson’s rule, one has to divide it into
an even number of blocks with the uniform width h. Each of the segments is
approximated by a parabola (instead of straight line as for trapezium rule).
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Example
p /2
ò
cos(x) dx
0
Number of
intervals
Trapezium rule
Simpson’s rule
10
1.183
1.1895
20
1.1932
1.1953
50
1.19695
1.19746
100
1.19772
1.19790
200
1.19799
1.19806
500
1.19810
1.19812
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