Mathematics - Area Under Curves Question with Solution | TestHub

$f\left(x\right)=e^{x-1}-e^{-\left|x-1\right|}=\left\{\begin{array}{ll}0,& x\le 1\\ e^{x-1}-e^{1-x},& x>1\end{array}\right.$ and

$g\left(x\right)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)$

At $x=0, f\left(0\right)=0 \& g\left(0\right)=\frac{e+e^{-1}}{2}$

At $x=1, f\left(1\right)=0 \& g\left(1\right)=\frac{1}{2}\left(e^{1-1}+e^{1-1}\right)=1$

For point of intersection of the two curves for $x>1$, $f\left(x\right)=g\left(x\right)$

$\Rightarrow e^{x-1}-e^{1-x}=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)$

$\Rightarrow 2e^{x-1}-2e^{1-x}=e^{x-1}+e^{1-x}$

$\Rightarrow e^{x-1}=3e^{1-x}$

$\Rightarrow {\left(e^{x-1}\right)}^2=3$

$\Rightarrow x=1+\ln \sqrt{3}$

$\therefore$Required area $=$ Area of the shaded region

$={\int }_0^{1}g\left(x\right)dx+{\int }_1^{1+\ln \sqrt{3}}\left\{g\left(x\right)-f\left(x\right)\right\}dx$

$={\int }_0^1\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)dx+{\int }_1^{1+\ln \sqrt{3}}\left\{\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)-\left(e^{x-1}-e^{1-x}\right)\right\}dx$

$=\frac{1}{2}{\int }_0^1\left(e^{x-1}+e^{1-x}\right)dx+\frac{1}{2}{\int }_1^{1+\ln \sqrt{3}}\left(3e^{1-x}-e^{x-1}\right)dx$

$=\frac{1}{2}{\left[e^{x-1}-e^{1-x}\right]}_0^1-\frac{1}{2}{\left[3e^{1-x}+e^{x-1}\right]}_1^{1+\ln \sqrt{3}}$

$=\frac{1}{2}\left(e-e^{-1}\right)-\frac{1}{2}\left[2\sqrt{3}-4\right]$

$=\frac{e-e^{-1}}{2}+\left(2-\sqrt{3}\right)$