Evaluate the integral of (e^-x+e^x)² dx from 0 to 1
Solution.
∫01(e−x+ex)2dx=∫01(e−2x+2e−xex+e2x)dx==(−12e−2x+2x+12e2x)∣01=−12e−2+2+12e2+12−12==−12e2+e22+2.\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}+2.0∫1(e−x+ex)2dx=0∫1(e−2x+2e−xex+e2x)dx==(−21e−2x+2x+21e2x)∣01=−21e−2+2+21e2+21−21==−2e21+2e2+2.
Answer. −12e2+e22+2.-\frac{1}{2e^2}+\frac{e^2}{2}+2.−2e21+2e2+2.
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