Answer
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, $$A=\int^1_0u^{1/2}du=\frac{2u^{3/2}}{3}\Big]^1_0$$ $$A=\frac{2}{3}(1-0)$$ $$A=\frac{2}{3}$$
