## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 5: Integrals - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises 5.6 - Page 303: 5

#### Answer

$a.\quad \displaystyle \frac{15}{16}$ $b.\quad$ 0

#### Work Step by Step

$a.$ Applying th.7, let $u=g(t)=1+t^{4}$ (a continuous function). Then, $\displaystyle \quad du=g'(t)dt=4t^{3}dt,\qquad t^{3}dt=\frac{du}{4}$ and, the borders change to $g(0)=1$ and $g(1)=2$. $\displaystyle \int_{0}^{1}t^{3}(1+t^{4})^{3}dt=\int_{1}^{2}u^{3}\cdot\frac{1}{4}du$ $=\displaystyle \frac{1}{4}\left[ \frac{u^{3+1}}{3+1} \right]_{1}^{2}$ $=\displaystyle \frac{1}{4}\left[ \frac{u^{4}}{4} \right]_{1}^{2}$ $=\displaystyle \frac{1}{16}(2^{4}-1^{4})$ $=\displaystyle \frac{15}{16}$ $b.$ Applying th.7, let $u=g(t)=1+t^{4}$ (a continuous function). Then, $\displaystyle \quad du=g'(t)dt=4t^{3}dt,\qquad t^{3}dt=\frac{du}{4}$ and, the borders change to $g(-1)=2$ and $g(1)=2$. $\displaystyle \int_{-1}^{1}t^{3}(1+t^{4})^{3}dt=\int_{2}^{2}u^{3}\cdot\frac{1}{4}du$ $=0$ (zero width integral)

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