Answer
$a)\frac{14}{3}$
b)$\frac{2}{3}$
Work Step by Step
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
$\int _{-1}^0\sqrt{y+1}dy$
=$\int ^1_0u^{1/2}du$
=$[\frac{2}{3}u^{3/2}]^1_0$
=$(\frac{2}{3})(1)^{3/2}-0$
=$\frac{2}{3}$
