If the area of the region (x,y):ax^2 y (1)/(x),1 x 2,0

Question

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 :

Accepted Answer

$(x,y):\frac{a}{{x}^{2}}\le y\le \frac{1}{x},1\le x\le 2,0$\Rightarrow \int \_{1}\^{2}(\frac{1}{x}-\frac{a}{{x}\^{2}})dx={|\ln |x|+\frac{a}{x}|}\_{2}\^{1}$

$$$\begin{gathered} (\ln 2+\frac{a}{2})-(\ln 1+a)=\ln 2-\frac{a}{2} \ \Rightarrow \, \frac{a}{2}=\frac{1}{7} \ \Rightarrow \, a=\frac{2}{7} \ \, 7a-3=\frac{2}{7}\times 7-3=-1 \end{gathered}$$

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