Question

The minimum value of the function $f(x)=\int\limits_0^2 e^{|x-t|} d t$ is :

• A. 2
• B. $2(e-1)$
• C. $2 e-1$
• D. $e(e-1)$

Answer 1

Answer: (B)

For $x \leq 0$
$f(x)=\int\limits_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right)$
For $0 < x < 2$
$f(x)=\int\limits_0^x e^{x-t} d t+\int_x^2 e^{t-x} d t=e^x+e^{2-x}-2$
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 $2(e-1)$