Answer
$$\int^{\sqrt7}_0t(t^2+1)^{1/3}dt=\frac{45}{8}$$
$$\int_{-\sqrt7}^0t(t^2+1)^{1/3}dt=-\frac{45}{8}$$
Work Step by Step
Part a) $$A=\int^{\sqrt7}_0t(t^2+1)^{1/3}dt$$
Set $u=t^2+1$, we then have $$du=2tdt$$ $$tdt=\frac{1}{2}du$$
For $t=0$, we have $u=1$ and for $t=\sqrt7$, we have $u=8$.
Therefore, $$A=\frac{1}{2}\int^{8}_1u^{1/3}du$$ $$A=\frac{1}{2}\times\frac{3u^{4/3}}{4}\Big]^{8}_1=\frac{3u^{4/3}}{8}\Big]^{8}_1$$ $$A=\frac{3}{8}(8^{4/3}-1)$$ $$A=\frac{3}{8}(16-1)=\frac{45}{8}$$
Part b) $$A=\int_{-\sqrt7}^0t(t^2+1)^{1/3}dt$$
We carry out a similar substitution as in part a).
For $t=0$, we have $u=1$ and for $t=-\sqrt7$, we have $u=8$.
Therefore, $$A=\frac{1}{2}\int^{1}_8u^{1/3}du$$ $$A=-\frac{1}{2}\int^{8}_1u^{1/3}du=-\frac{45}{8}$$
