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  • The minimum value of the function <img alt="f(x)=\int_{0}^{2} e^{|x-t|} d t" src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x%29%3D%5Cint_%7B0%7D%5E%7B2%7D%20e%5E%7B%7Cx-t%7C%7D%20d%20t"

The minimum value of the function f(x)=\int_{0}^{2} e^{|x-t|} d t is:

Option: 1

e(e-1)


Option: 2

2(e-1)


Option: 3

2


Option: 4

2 e-1


Answers (1)

\text{Case I} \quad \quad\mathrm{x}<0
     f(x)=\int_{0}^{2} e^{-(x-t)} d t
=e^{-x} \int_{0}^{2} e^{t} d t=e^{-x}\left(e^{2}-1\right)

Case II
(0<x<2) \quad f(x)=\int_{0}^{x} e^{x-t} d t+\int_{x}^{2} e^{-(x-t)} d t
                                 =\mathrm{e}^{\mathrm{x}}\left(-\mathrm{e}^{-\mathrm{t}}\right)_{0}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\left[\mathrm{e}^{\mathrm{t}}\right]_{\mathrm{x}}^{2}
                                  =\mathrm{e}^{\mathrm{x}}\left[-\mathrm{e}^{-\mathrm{x}}+1\right]+\mathrm{e}^{-\mathrm{x}}\left[\mathrm{e}^{2}-\mathrm{e}^{\mathrm{x}}\right]
                                 =-1+\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{2-\mathrm{x}}-1

   
Case III
\mathrm{x} \geq 2 \quad \mathrm{f}(\mathrm{x})=\int_{0}^{2} \mathrm{e}^{(\mathrm{x}-\mathrm{t})} \mathrm{dt}
                 =\mathrm{e}^{\mathrm{x}}\left[-\mathrm{e}^{-\mathrm{t}}\right]_{0}^{2}
                 =\mathrm{e}^{\mathrm{x}}\left[-\mathrm{e}^{-2}+1\right]
                 =\mathrm{e}^{\mathrm{x}}\left(1-\mathrm{e}^{-2}\right)

\mathrm{f}(\mathrm{x})=\left[\begin{array}{ccc} \mathrm{e}^{-\mathrm{x}}\left(\mathrm{e}^{2}-1\right), & \mathrm{x} \leq 0 \rightarrow & \left(\mathrm{e}^{2}-1\right) \\ \mathrm{e}^{\mathrm{x}}+\mathrm{e}^{2-\mathrm{x}}-2, & 0 \leq \mathrm{x} \leq 2 \rightarrow & 2(\mathrm{e}-1) \\ \operatorname{ex}\left(1-\mathrm{e}^{-2}\right), & \mathrm{x} \geq 2 \rightarrow & \left(\mathrm{e}^{2}-1\right) \end{array}\right.

Minimum value =2(\mathrm{e}-1)
 

Posted by

Ramraj Saini

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