   Chapter 7.6, Problem 10E

Chapter
Section
Textbook Problem

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

To determine

To evaluate the integral 2y23y2dy

Explanation

Calculation: Given 2y23y2dy

Substitute 2y=u,then2dy=du

2y23y2dy=u23(u2)2du2=2u23u2du

From Table of Integrals, Entry No 24 gives

u2a2u2du=u2a2u+ln(u+u2a2)+C

Therefore

2y23

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