If the area of the region {(x, y): frac{a}{x^ | vedclass

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(C) The area of the region is given by the integral: $Area = \int_1^2 \left(\frac{1}{x} - \frac{a}{x^2}\right) dx$ $= \left[ \ln|x| + \frac{a}{x} \rig

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$-1$

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(C) The area of the region is given by the integral: $Area = \int_1^2 \left(\frac{1}{x} - \frac{a}{x^2}\right) dx$ $= \left[ \ln|x| + \frac{a}{x} \right]_1^2$ $= (\ln 2 + \frac{a}{2}) - (\ln 1 + \frac{a}{1})$ $= \ln 2 + \frac{a}{2} - a = \ln 2 - \frac{a}{2}$ Given that the area is $(\ln 2) - \frac{1}{7}$,we equate: $\ln 2 - \frac{a}{2} = \ln 2 - \frac{1}{7}$ $-\frac{a}{2} = -\frac{1}{7}$ $a = \frac{2}{7}$ Now,we calculate $7a - 3$: $7(\frac{2}{7}) - 3 = 2 - 3 = -1$

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

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