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91Ó°ÊÓ

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

∫01x1−xdx

Short Answer

Expert verified

The solution of the given integral is ∫01x1−xdx=23.

Step by step solution

01

Step 1. Given Information 

Solving the given integrals.

∫01x1−xdx

02

Step 2. Using the substitution method. 

Let

u=1−xdudx=−1du=−dx−du=dx

03

Step 3. We will now write the limits of integration  in terms of the new variable u.

When x=0, we have

u=1−xu=1−0u=1

When role="math" localid="1649087278747" x=1, we have

u=1−xu=1−1u=0

04

Step 4. Using the information in equations, we can change variables completely:

∫01x1−xdx=−∫10udu∫01x1−xdx=−∫10u1/2du∫01x1−xdx=−u1/2+11/2+110∫01x1−xdx=−u3/23/210∫01x1−xdx=−23u3/210∫01x1−xdx=−23(0)3/2-(1)3/2∫01x1−xdx=−230-1∫01x1−xdx=−23(-1)∫01x1−xdx=23

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