Question

The area bounded by $x-$axis, the curve $y=(1+\frac{8}{x^2})$ and the ordinates at $x=2$ and $x=4$ is divided into two equal parts at $x=a$. Then $a^2-\sqrt{2}a-2=$

Answer 1

We have,
$y=1+\frac{8}{x^2}$
Required area$=\underset{2}{\overset{4}{\int }}ydx$
$=\underset{2}{\overset{4}{\int }}(1+\frac{8}{x^2})dx={[x-\frac{8}{x}]}_2^4=4$
If $x=a$ bisects the area then we have
$\underset{2}{\overset{a}{\int }}(1+\frac{8}{x^2})dx={[x-\frac{8}{x}]}_2^a=[a-\frac{8}{a}-2+4]=\frac{4}{2}$
$\Rightarrow a-\frac{8}{a}=0\Rightarrow a^2=8$
$\Rightarrow a=\pm 2\sqrt{2}$
Since $2<a<4$
$\therefore a=2\sqrt{2}\Rightarrow a^2-\sqrt{2}a-2=2$