Animated Solution for Mathematics - Definite Integration: If the area of the region $\{(x,y):\frac{a}{x^2}\le y\le \frac{1}{x},1\le x\le 2,0<a<1\}$ is $({\log }_{e}2)-\frac{1}{7}$ then the value of $7a-3$ is equal to:

Welcome, fellow traveler of the JEE path. Today, we are peeling back the layers of a geometric puzzle defined by the region $\{(x,y):\frac{a}{x^2}\le y\le \frac{1}{x},1\le x\le 2,0<a<1\}$.

Imagine the Cartesian plane with two curves. The first is $y=\frac{1}{x}$, a hyperbola that descends as $x$ increases. The second is $y=\frac{a}{x^2}$, which remains below the hyperbola since $0<a<1$.

We are restricted to the corridor between $x=1$ and $x=2$. In this region, the hyperbola $y=\frac{1}{x}$ acts as the ceiling, and $y=\frac{a}{x^2}$ acts as the floor.

To find the area of the region trapped between these curves, we sum an infinite number of vertical strips. Each strip has a height equal to the difference between the ceiling and the floor: $(\frac{1}{x}-\frac{a}{x^2})$.

Calculus allows us to translate this geometric intuition into a definite integral. The area $A$ is defined as:

Integration is linear, allowing us to handle the terms separately. First, we evaluate the integral of the first term:

Since $\ln 1=0$, this term simplifies cleanly to $\ln 2$.

Next, we tackle the second term: ${\int }_1^2-\frac{a}{x^2}dx$. Pulling the constant $-a$ outside, we apply the power rule $\int x^{n}dx=\frac{x^{n+1}}{n+1}$:

Evaluating this expression at the boundaries gives:

Combining these results, our total area $A$ is:

We equate this to the given area value of $\ln 2-\frac{1}{7}$:

The $\ln 2$ terms cancel out, leaving us with:

The problem asks for the value of $7a-3$. Substituting our value of $a$:

The final answer is $-1$.