The area enclosed by the curves [y= _e(x+e^2), x= _e((2)/(y))] and [x= _e 2], above the li…

Question

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

Accepted Answer

According to NTA, the required region $A_2$ which is shaded in crossed lines and comes out to be

$$
A_2=\int_1^2\left(\ln \frac{2}{y}-e^y+e^2\right) d y=1+e-\ln 2
$$

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

$$
\begin{aligned}
A_1 &=\int_0^{\ln 2}\left(\ln \left(x+e^2\right)-2 e^{-x}\right) d x \\
&=\left.\left\{\left(x+e^2\right) \ln \left(x+e^2\right)-x+2 e^{-x}\right\}\right|_0 ^{\ln 2} \\
&=\left(\ln 2+e^2\right) \ln \left(\ln 2+e^2\right)-\ln 2+1 \\
& \quad-2 e^2-2 \\
&=\left(\ln 2+e^2\right) \ln \left(\ln 2+e^2\right)-\ln 2-2 e^2-1
\end{aligned}
$$

Not given in any option.

The region asked in the question is bounded by three curves

$$
\begin{aligned}
&y=\ln \left(x+e^2\right) \\
&x=\ln \left(\frac{2}{y}\right) \\
&x=\ln 2
\end{aligned}
$$

There is only one region which satisfies above requirement and which also lies above line $y=1$

Line $y=1$ may or may not be the boundary of the region.

Solutions · 1