Question

${\int }_a^b{x}^{3}dx=0$ and ${\int }_a^b{x}^{2}dx=\frac{2}{3};$ Find both ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$.

Answer 1

${\int }_a^b{x}^{3}dx=0$

$\Rightarrow b^4-a^4=0=(a^2-b^2)(a^2+b^2)=0$

$\Rightarrow a^2=b^2$ [$\because$ $a^2\ne {-b}^2$]

$\Rightarrow a=b$ or $a=-b$

${\int }_a^b{x}^{2}dx=2/3$

$\Rightarrow b^3-a^3=2=(b-a)(b^2+ab+a^2)$

We know, either $a=b$ or $a=-b$

But $a\ne b$ [$\because$ $b^3-a^3\ne 0$]

$\therefore$ $a=-b$

$\Rightarrow (b-a)(b^2+ab+a^2)=2a(a^2+a^2+a^2)$

$\Rightarrow {6a}^3=2$

$\Rightarrow a^3=\frac{1}{3}$

$\therefore$ $a=\frac{1}{\sqrt[3]{33}},b=\frac{-1}{\sqrt[3]{33}}$