Question
Asked Nov 23, 2019
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Evaluate the integral.
In 5
2х
dx
Зе
0
In 5
| Зе 2X dx 3D
0
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Evaluate the integral. In 5 2х dx Зе 0 In 5 | Зе 2X dx 3D 0

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Expert Answer

Step 1

To evaluate the below integral

x=In(5)
3e2 dx
Jx=O
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x=In(5) 3e2 dx Jx=O

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Step 2

Evaluating the given integral using the substituting method as shown.

x=In{5)
3e2 dx
Let u 2x
x=u/2
du /2{On differentiating both sides
dr
Also at x 0,u = 0 and at x = In (5),u = 2ln(5
Substitute all these values in the given integral, it gives
1
21n(5)
edu
2
x=In(5)
3e2 dx 3
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x=In{5) 3e2 dx Let u 2x x=u/2 du /2{On differentiating both sides dr Also at x 0,u = 0 and at x = In (5),u = 2ln(5 Substitute all these values in the given integral, it gives 1 21n(5) edu 2 x=In(5) 3e2 dx 3

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Step 3

Solve as shown...

3
2 In(5)
2 In( 5)
3-
{fe'dr = e
edu
2
du
3
e"
10
2 Im(5)
е
-e)
_
2
3
In(5)
{alnx=Inx
1nx
= X
2(25-1)
3
24
2
= 36
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3 2 In(5) 2 In( 5) 3- {fe'dr = e edu 2 du 3 e" 10 2 Im(5) е -e) _ 2 3 In(5) {alnx=Inx 1nx = X 2(25-1) 3 24 2 = 36

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Tagged in

Math

Calculus

Integration