### Definite Integration Question 5

#### Question 5 - 25 January - Shift 1

The minimum value of the function $f(x)=\int_0^{2} e^{|x-t|} d t$ is

(1) $2(e-1)$

(2) $2 e-1$

(3) 2

(4) $e(e-1)$

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Properties of definite integral , Properties of modulus function, Condition for increasing and decreasing function, Relation between means , Integration of exponential function

For $x \leq 0$

$f(x)=\int_0^{2} e^{t-x} d t=e^{-x}(e^{2}-1)$

For $0<x<2$

$f(x)=\int_0^{x} e^{x-t} d t+\int_x^{2} e^{t-x} d t=e^{x t}+e^{2-x}-2$

For $x \geq 2$

$f(x)=\int_0^{2} e^{x-t} d t=e^{x-2}(e^{2}-1)$

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 $2(e-1)$