## Thomas' Calculus 13th Edition

$a)\frac{14}{3}$ b)$\frac{2}{3}$
a)Let u=y+1=>du=dy; y=0=>u=1, y=3=>u=4 $\int _{0}^{3}\sqrt{y+1}dy$ =$\int_1^4u^{1/2}du$ =$[\frac{2}{3}u^{3/2}]^4_1$ =$(\frac{2}{3})(4)^{3/2}-(\frac{2}{3})(1)^{3/2}$ =$\frac{2}{3}(8)-(\frac{2}{3})(1)$ =$\frac{14}{3}$ --- b)Use the same substitution for u as in part (a); y=-1=>u=0, y=0=>u=1 $\int _{-1}^0\sqrt{y+1}dy$ =$\int ^1_0u^{1/2}du$ =$[\frac{2}{3}u^{3/2}]^1_0$ =$(\frac{2}{3})(1)^{3/2}-0$ =$\frac{2}{3}$