   Chapter 7.6, Problem 22E

Chapter
Section
Textbook Problem

# Use the Table of Integrals on Reference Pages 6–10 to evaluate the integral. ∫ 0 2 x 3 4 x 2 − x 4 d x

To determine

To Find: The value of the integral 02x34x2x4dx

Explanation

Calculations: Given 02x34x2x4dx

02x34x2x4dx=02x24x2(x2)2xdx

=04u4uu2.12du

=1204u4(224u+u2)du

=1204u22(u2)2du

Substituted u=x2x=u12

du=ddx(x2)dx=2xdx

xdx=12du

When x=0u=0

When x=2u=4

Now making another substitution y = u-2

02x34x2x4dx=1222(y+2)22y2dy

=1222y22y2dy+1222222y2dy

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