If the area of the region $\left\{(x, y): \frac{\mathrm{a}}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0

Question

If the area of the region $\left\{(x, y): \frac{\mathrm{a}}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0 < \mathrm{a} < 1\right\}$ is $\left(\log _{\mathrm{e}} 2\right)-\frac{1}{7}$ then the value of $7 \mathrm{a}-3$ is equal to:

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { area } \int_1^2\left(\frac{1}{x}-\frac{a}{x^2}ight) d x \ & {\left[\ell n x+\frac{a}{x}ight]_1^2} \ & \ell \operatorname{nn} 2+\frac{a}{2}-a=\log _e 2-\frac{1}{7} \ & \frac{-a}{2}=-\frac{1}{7} \ & a=\frac{2}{7} \ & 7 a=2 \ & 7 a-3=-1\end{aligned}$

If the area of the region $\left\{(x, y): \frac{\mathrm{a}}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0…

Solution