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: 1


a) $$\int^3_0\sqrt{y+1}dy=\frac{14}{3}$$ b) $$\int^0_{-1}\sqrt{y+1}dy=\frac{2}{3}$$

Work Step by Step

Part a) $$A=\int^3_0\sqrt{y+1}dy$$ Set $u=y+1$, we then have $$du=dy$$ For $y=0$, we have $u=1$ and for $y=3$, we have $u=4$. Therefore, $$A=\int^4_1\sqrt udu=\int^4_1u^{1/2}du$$ $$A=\frac{2u^{3/2}}{3}\Big]^4_1$$ $$A=\frac{2}{3}(4^{3/2}-1)=\frac{2}{3}(8-1)$$ $$A=\frac{14}{3}$$ Part b) $$A=\int^0_{-1}\sqrt{y+1}dy$$ We carry out a similar substitution as in part a). For $y=0$, we have $u=1$ and for $y=-1$, we have $u=0$. Therefore, $$A=\int^1_0u^{1/2}du=\frac{2u^{3/2}}{3}\Big]^1_0$$ $$A=\frac{2}{3}(1-0)$$ $$A=\frac{2}{3}$$
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