The minimum value of the function $f(x) = \int\limits_0^2 {{e^{|x - t|}}dt}$ is :
Solution
$f(x)=\int_0^2 e^{|x-t|} d t$<br/><br/>
For $x>2$<br/><br/>
$f(x)=\int_0^2 e^{x-t} d t=e^x\left(1-e^{-2}\right)$<br/><br/>
For $x<0$<br/><br/>
$f(x)=\int_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right)$<br/><br/>
For $x \in[0,2]$<br/><br/>
$$
\begin{aligned}
& f(x)=\int_0^x e^{x-t} d t + \int_x^2 e^{t-x} d t \\\\
& =e^{2-x}+e^x-2
\end{aligned}
$$<br/><br/>
For $x>2$<br/><br/>
$\left.f(x)\right|_{\min =e^2-1}$<br/><br/>
For $\mathrm{x}<0$<br/><br/>
$\left.f(x)\right|_{\min =e^2-1}$<br/><br/>
For $x \in[0,2]$<br/><br/>
$\left.f(x)\right|_{\min }=2(e-1)$
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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