Question

Evaluate the integral of (e^-x+e^x)² dx from 0 to 1

Answer 1

Solution.

$\int\limits_0^1 (e^{-x}+e^x)^2dx=\int\limits_0^1(e^{-2x}+2e^{-x}e^x+e^{2x})dx=\newline =(-\frac{1}{2}e^{-2x}+2x+\frac{1}{2}e^{2x})|_0^1=-\frac{1}{2}e^{-2}+2+\frac{1}{2}e^2+\frac{1}{2} -\frac{1}{2}=\newline =-\frac{1}{2e^2}+\frac{e^2}{2}+2.$

Answer. $-\frac{1}{2e^2}+\frac{e^2}{2}+2.$