Explanation:
(B) : Given,
\(\int_{a}^{b} x^{3} d x=0\)
\({\left[\frac{x^{4}}{4}\right]_{a}^{b}=0}\)
\(\frac{b^{4}-a^{4}}{4}=0\)
\(b^{4}-a^{4}=0\)
\(b^{4}=a^{4}\)
\(b= \pm a \tag{i}\)
and, \(\quad \int_{a}^{b} x^{2} d x=\frac{2}{3}\)
\(\left[\frac{\mathrm{x}^{3}}{3}\right]_{\mathrm{a}}^{\mathrm{b}}=\frac{2}{3}\)
\(\frac{\mathrm{b}^{3}-\mathrm{a}^{3}}{3}=\frac{2}{3}\)
\(b^{3}-a^{3}=2\)
Let, b = a
Then, \(\quad b^3-a^3=a^3-a^3=0 \neq 2\)
Now, let \(\mathrm{b}=-\mathrm{a}\)
Then, \(\quad b^\beta-a^3\)
\(\Rightarrow (-a)^3-a^3=2\)
\(\Rightarrow \quad -2 a^3=2\)
\(\Rightarrow \quad a^3=-1\)
\(a=-1\)
\(b=-a=-(-1)=1\)
So,\((\mathrm{a}, \mathrm{b})=(-1,1)\)