If the line $x = a$ bisects the area under the curve $y=\frac{1}{x^2}, 1≤x≤9$, then a is equal to

$\frac{4}{9}$ $\frac{9}{5}$ $\frac{5}{9}$ $\frac{9}{4}$

$\frac{9}{5}$

It is given that $2\int\limits_1^a\frac{1}{x^2}dx=\int\limits_1^9\frac{1}{x^2}dx$ $⇒2\left[-\frac{1}{x}\right]_1^a=\left[-\frac{1}{x}\right]_1^9$ $⇒2\left(-\frac{1}{a}+1\right)=\left(-\frac{1}{9}+1\right)$ $⇒-\frac{2}{a}+2=-\frac{1}{9}+1⇒-\frac{2}{a}=-\frac{10}{9}⇒a=\frac{9}{5}$