If the area of the region $\{(x,y):-1\le x\le1,\ 0\le y\le a+e^{|x|}-e^{-x},\ a0\}$ is $\frac{e^2+8e+1}e$, then the value of $a$ is :

Question

If the area of the region $\{(x,y):-1\le x\le1,\ 0\le y\le a+e^{|x|}-e^{-x},\ a>0\}$ is $\frac{e^2+8e+1}e$, then the value of $a$ is :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Area $=\int_{-1}^1\left(a+e^{|x|}-e^{-x}\right)dx=2a+2(e-1)-\left(e-\frac1e\right)=2a+e-2+\frac1e$. Setting this equal to $e+8+\frac1e$ gives $2a-2=8$, so $a=5$. Answer (3).

Q. If the area of the region $\{(x,y):-1\le x\le1,\ 0\le y\le a+e^{|x|}-e^{-x},\ a>0\}$ is $\frac{e^2+8e+1}e$, then the value of $a$ is :

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?