The area bounded by the x-axis; part of the curve $y=(1+\frac{8}{x^2})$ and the ordinates at x = 2 and x = 4. If the ordinate at x = a divides the area into two parts, then the value of ‘a’ is:

$2\sqrt{2}$

$A=\int_2^4y\,dx=\int_2^4(1+\frac{8}{x^2})dx=[x-\frac{8}{x}]_2^4=4$

Also $A_1=\int_2^ay\,dx=\frac{1}{2}A=2[as[x-\frac{8}{x}]_2^a=2]$

or $(a-2)-8(\frac{1}{a}-\frac{1}{2})=2$ or $a-\frac{8}{a}=0$

$∴a^2-8=0$

$∴a=2\sqrt{2}$