JEE Main 2023 — Definite Integration Question with Solution

Given,

$f\left(x\right)={\int }_0^2{e}^{\left|x-t\right|}dt$

Now For $x\le 0$

$f\left(x\right)={\int }_0^2{e}^{t-x}dt=e^{-x}\left(e^2-1\right)$

And for $0<x<2$

$f\left(x\right)={\int }_0^x{e}^{x-t}dt+{\int }_x^2{e}^{t-x}dt=e^x+e^{2-x}-2$

For $\mathrm{x}\ge 2$

$f\left(x\right)={\int }_0^2{e}^{x-t}dt=e^{x-2}\left(e^2-1\right)$

Now for $x\le 0, f\left(x\right)$ is decreasing as $e^{-x}$ is decreasing function and $x\ge 2, f\left(x\right)$ is increasing as $e^{x-2}$ is increasing function,

So, minimum value of $f\left(x\right)$ lies in $x\in \left(0,2\right)$

Applying A.M $\ge$ G.M in $e^x+e^{2-x}$we get,

$\frac{e^x+e^{2-x}}{2}\ge \sqrt{e^x\times e^{2-x}}$

$\Rightarrow e^x+e^{2-x}\ge 2e$

Hence, the minimum value of $f\left(x\right)$ is $2e-2=2\left(e-1\right)$