The area between the curves y = xex and y = xe–x and the line x = 1 is

The line x = 1 meets the curves in A(1, e) and B(1, 1/e). Both the curves pass through the origin.

The required area

$A=\int\limits_0^1(y_1-y_2)dx=\int\limits_0^1(xe^x-xe^{-x})dx$

$=[x\{e^x+e^{-x}\}]_0^1-\int\limits_0^1(e^x+e^{-x}).1dx$

$=(e+\frac{1}{e})-[e^x+e^{-x}]_0^1=(e+\frac{1}{e})-(e-\frac{1}{e})=\frac{2}{e}$ sq. units.

Hence (C) is the correct answer.