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: 15



Work Step by Step

$$A=\int^1_0\sqrt{t^5+2t}(5t^4+2)dt$$ We set $u=t^5+2t$, which means $$du=(5t^4+2)dt$$ For $t=0$, we have $u=0^5+2\times0=0$ For $t=1$, we have $u=1^5+2\times1=3$ Therefore, $$A=\int^3_0\sqrt udu=\int^3_0u^{1/2}du$$ $$A=\frac{2}{3}u^{3/2}\Big]^3_0=\frac{2}{3}\times3^{3/2}=2\times3^{1/2}$$ $$A=2\sqrt3$$
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