Mathematics - Area Under Curves Question with Solution | TestHub

The area enclosed by the curves $y={\log }_e\left(x+e^2\right), x={\log }_e\left(\frac{2}{y}\right)$ and $x={\log }_{e}2$, above the line $y=1$ is

Now plotting the diagram of given functions we get,

According to JEE Mains, we have to calculate the required region $A_2$ which is shaded in crossed lines and comes out to be

$A_2={\int }_1^2\left(\ln \frac{2}{y}-e^y+e^2\right)dy=1+e-\ln 2$

But according to the question the required region $A_1$ comes out to be shaded in parallel lines, which can be obtained as

$A_1={\int }_0^{\ln 2}\left(\ln \left(x+e^2\right)-2e^{-x}\right)dx$

${\overline{)=\left\{\left(x+e^2\right)\ln \left(x+e^2\right)-x+2e^{-x}\right\}}}_0^{\text{ln}2}$

$=\left(\ln 2+e^2\right)\ln \left(\ln 2+e^2\right)-\ln 2+1-2e^2-2$

$=\left(\ln 2+e^2\right)\ln \left(\ln 2+e^2\right)-\ln 2-2e^2-1$

Not given in any option.