Discussion
The value of integral \(\int_{-2}^{+2} x\, e^{-2x^{2}}\, dx\) is
Answer: A
The integral equals 0.
The integrand \(f(x) = x\,e^{-2x^{2}}\) is odd: \(f(-x) = -x\,e^{-2x^{2}} = -f(x)\). The integral of an odd function over limits symmetric about zero vanishes, because the area for negative x exactly cancels the area for positive x.
Directly, with \(u = -2x^{2}\):
\[ \int x\,e^{-2x^{2}}\,dx = -\tfrac{1}{4}e^{-2x^{2}} \]
which takes the same value at x = +2 and x = −2, so the difference is zero. The value 1/2 would come from integrating only from 0 to infinity and doubling, which ignores the sign of x.
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