I need to solve an integral: $\int_{0}^{\infty}|x-c|e^{-2x}dx$ I got solution for this which is right ,(by splitting the integral) .I got solution for $c$ if negative as $\frac{1}{4}-\frac{c}{2}$ and for $c$ if positive $\frac{e^{-2c}}{2} - \frac{1}{4} + \frac{c}{2}$ and now i need to solve for $c \in$ $\R$ for which integral is minimum. I don't understand the question :( can someone give a hint?
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$\int_{0}^{\infty}|x-c|e^{-2x}dx$ is a function of $c$ : so you can find $f'=0 \\f(c)=\int_{0}^{\infty}|x-c|e^{-2x}dx$ $$f(c)=\int_{0}^{c}|x-c|e^{-2x}dx+\int_{c}^{\infty}|x-c|e^{-2x}dx=\\ \int_{0}^{c}(-(x-c))e^{-2x}dx+\int_{c}^{\infty}(x-c)e^{-2x}dx=\\ (-(x-c))\frac{e^{-2x}}{-2}-(\frac{e^{-2x}}{-4}) +((x-c)\frac{e^{-2x}}{-2}+(\frac{e^{-2x}}{-4}) \\ (0+\frac{e^{-2c}}{2}+\frac{c}{2}-\frac14)+(0+0-(0+\frac{e^{-2c}}{-4})\\ \frac14(3e^{-2c}+2c-1)$$ now find $f'(c)$ $$f'=\frac14(-6e^{-2c}+2)=0 \\ \to e^{-2c}=\frac26 $$now find the root of $f'=0$ $$\\-2c=ln \frac13\\ \to \\c=\frac12\ln3$$ finally check for minimum $$f''(\frac12 \ln3)=\frac14(+12e^{-2c}+0)>0$$ ,this point is minimum