Question

If the ordinate $x=a$ divides the area bounded by x=axis, part of the curve $y=1+\frac{8}{x^2}$ and the ordinates $x=2,x=4$, into two equal parts, then $a$ is equal to

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $3\sqrt{2}$
• D. None of these

Answer 1

The correct option is B $2\sqrt{2}$

The area bounded by the curve $y=1+\frac{8}{x^2}$, x-axis and the ordinates $x=2,x=4$ is
$={\int }_2^{4}ydx={\int }_2^4(1+\frac{8}{x^2})dx={[x-\frac{8}{x}]}_2^4=(4-2)-(2-4)=4$
Since $x=a$ divides this area into two equal parts,
$\therefore$ required area $=2{\int }_2^{a}ydx$
$\therefore 4=2{\int }_2^a(1+\frac{8}{x^2})dx\Rightarrow 2={[x-\frac{8}{x}]}_2^a$
$=(a-\frac{8}{a})-(2-4)\Rightarrow a^2=8\Rightarrow a=2\sqrt{2}$