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 $2(e-1)$
