7.2 definite integral — Maths STD 12 Science — Question

$f(x)=\int \limits_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right)$

For $0 < x < 2$

$f(x)=\int \limits_0^x e^{x-t} d t+\int \limits_x^2 e^{t-x} d t=e^x+e^{2-x}-2$

For $x \geq 2$

$f(x)=\int \limits_0^2 e^{x-t} d t=e^{x-2}\left(e^2-1\right)$

For $x \leq 0, f ( x )$ is $\downarrow$ and $x \geq 2, f ( x )$ is $\uparrow$

$\therefore$ Minimum value of $f ( x )$ lies in $x \in(0,2)$

Applying A.M $\geq$ $G.M$

minimum value of $f(x)$ is $2(e-1)$