Answer
$$\int^{1}_0t^3(1+t^4)^3dt=\frac{15}{16}$$
$$\int^{1}_{-1}t^3(1+t^4)^3dt=0$$
Work Step by Step
Part a) $$A=\int^{1}_0t^3(1+t^4)^3dt$$
Set $u=1+t^4$, we then have $$du=4t^3dt$$ $$t^3dt=\frac{1}{4}du$$
For $t=0$, we have $u=1$ and for $t=1$, we have $u=2$.
Therefore, $$A=\frac{1}{4}\int^{2}_1u^3du$$ $$A=\frac{1}{4}\times\frac{u^4}{4}\Big]^{2}_1=\frac{u^{4}}{16}\Big]^{2}_1$$ $$A=\frac{1}{16}(2^4-1)=\frac{15}{16}$$
Part b) $$A=\int^{1}_{-1}t^3(1+t^4)^3dt$$
We carry out a similar substitution as in part a).
For $t=-1$, we have $u=2$ and for $t=1$, we have $u=2$.
Therefore, $$A=\frac{1}{4}\int^{2}_{2}3u^3du=0$$
