University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 337: 5


$$\int^{1}_0t^3(1+t^4)^3dt=\frac{15}{16}$$ $$\int^{1}_{-1}t^3(1+t^4)^3dt=0$$

Work Step by Step

Part a) $$A=\int^{1}_0t^3(1+t^4)^3dt$$ Set $u=1+t^4$, we then have $$du=4t^3dt$$ $$t^3dt=\frac{1}{4}du$$ For $t=0$, we have $u=1$ and for $t=1$, we have $u=2$. Therefore, $$A=\frac{1}{4}\int^{2}_1u^3du$$ $$A=\frac{1}{4}\times\frac{u^4}{4}\Big]^{2}_1=\frac{u^{4}}{16}\Big]^{2}_1$$ $$A=\frac{1}{16}(2^4-1)=\frac{15}{16}$$ Part b) $$A=\int^{1}_{-1}t^3(1+t^4)^3dt$$ We carry out a similar substitution as in part a). For $t=-1$, we have $u=2$ and for $t=1$, we have $u=2$. Therefore, $$A=\frac{1}{4}\int^{2}_{2}3u^3du=0$$
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