   Chapter 7.3, Problem 24E

Chapter
Section
Textbook Problem

# Evaluate the integral. ∫ 0 1 x − x 2 d x

To determine

To evaluate: The given integral 01xx2dx.

Explanation

Integration involving terms of the form a2x2 can be simplified by using the trigonometric substitution x=asinθ.

Formula used:

The identity, cos2x=1sin2x

Given:

The integral, 01xx2dx

Calculation:

The integral can be written as:

01xx2dx=01(x2x)dx=01(x2212x+1414)dx=01((x12)214)dx=0114(x12)2dx

Use the substitution u=x12, then du=dx and the limit of integration will be:

x0u12andx1u12

So, the integration in terms of u will be:

01xx2dx=121214u2du

Substitute for x as u=12sinθ. Take the derivative of the substitution term:

u=12sinθdu=12cosθdθ

Here, π2θπ2. The limits of integration will change as:

u1212sinθ=12sinθ=1θπ2andx1212sinθ=12sinθ=1θπ2

Substitute for u and du in the given integral to get:

01xx2dx=12

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