The minimum value of the function $f(x)=\int\limits_0^2 e^{|x-t|} d t$ is :
- 2
- $2(e-1)$
- $2 e-1$
- $e(e-1)$
The Correct Option is B
Solution and Explanation
For x≤0
f(x)=0∫2et−xdt=e−x(e2−1)
For 0<x<2
f(x)=0∫xex−tdt+∫x2et−xdt=ex+e2−x−2
For x≥2
f(x)=0∫2ex−tdt=ex−2(e2−1)
For x≤0,f(x) is ↓ and x≥2,f(x) is ↑
∴ Minimum value of f(x) lies in x∈(0,2)
Applying A.M≥G.M,
minimum value of f(x) is 2(e−1)
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
