If the area of the region \[ \left\{(x, y) : \frac{a}{x^2} \leq y \leq \frac{1}{x}, \, 1 \leq x \leq 2, \, 0<a<1 \right\} \] is \[ (\log 2) - \frac{1}{7}, \] then the value of $7a - 3$ is equal to:
The region's area is computed as: \[\text{Area} = \int_1^2 \left( \frac{1}{x} - \frac{a}{x^2} \right) dx\] The integral evaluation is: \[= \left[ \ln x + \frac{a}{x} \right]_1^2\] \[= \ln 2 + \frac{a}{2} - a = \log_2 2 - \frac{1}{7}\] By equating terms and solving for \( a \): \[-\frac{a}{2} = -\frac{1}{7}\] \[a = \frac{2}{7}\] The calculation of \( 7a - 3 \) follows: \[7a = 2\] \[7a - 3 = -1\]
