If ${\int }_a^b x^{3}dx=0$ and if ${\int }_a^b x^{2}dx=\frac{2}{3}$, then the values of a and b are respectively :

Given ${\int }_a^b x^{3}dx=0$

$\begin{array}{l}{\left.\Rightarrow \frac{x^4}{4}\right|}_a^b=0\\ \Rightarrow \frac{b^4}{4}-\frac{a^4}{4}=0\end{array}$

Thus $b^4-a^4=0$

$\Rightarrow \left(b^2-a^2\right)\left(b^2+a^2\right)=0$

$\Rightarrow (b-a)(b+a)\left(b^2+a^2\right)=0$

Either $b-\mathrm{a}=0\text{ or }\mathrm{b}+\mathrm{a}=0\text{ or }{\mathrm{b}}^2+{\mathrm{a}}^2=0$

But $b-a\ne 0$ and $a^2+b^2\ne 0$

$\therefore b+a=0\Rightarrow a=-b$

Now, also 

${\int }_a^b x^{2}dx={\left.\frac{x^3}{3}\right|}_a^b=\frac{b^3-a^3}{3}=\frac{2}{3}$

$\begin{array}{l}\Rightarrow (b-a)\left(b^2+a^2+ab\right)=2\\ \Rightarrow \left(2b\right)\left(b^2+b^2-b^2\right)=2\\ \Rightarrow 2b^3=2\Rightarrow b=1\end{array}$

So, $a=-1$, thus a and b are  -1, 1  respectively.