Evaluate the following: integral -infinity to +infinity xe^-x^2 dx.

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We are asked to evaluate `int_(-oo)^(oo)e^(-x^2)xdx`

I. The easy way is to recognize `xe^(-x^2)` as an odd function. ( f(-x)=-f(x) ). Since the limits of integration are symmetric about the origin, the integral is zero.

`int_(-oo)^(oo)e^(-x^2)xdx=0`

II. If you did not see that, we can apply what we know of improper integrals. Improper integrals are integrals with either or both limits at infinity(negative infinity) or an integrand that approaches infinity at one of the limits.

We can rewrite the improper integral as the limit of an integral. Note that since both limits are infinite, we have to write two limits. Thus:

`int_(-oo)^(oo)e^(-x^2)xdx=lim_(a ->-oo) int_(a)^(0) e^(-x^2)xdx+lim_(b->oo)int_(0)^(b)e^(-x^2)xdx`

To evaluate the first limit, let u=-x^2, du=-2xdx. Then we have:

`lim_(a->-oo)-1/2 int_(a)^(0)e^(u)du=-1/2e^(u)|_(-oo)^(0)=-1/2(1-0)=-1/2`

Using the same substitution for the second limit we get:

`lim_(b->oo)1/2int_(0)^(b)e^(u)du=-1/2e^(u)|_(0)^(oo)=-1/2(0-1)=1/2`

Summing the limits we get -1/2 + 1/2 = 0.

If either limit was unbounded, we could resort to a comparison test. In no way can we "evaluate" `-oo+oo` ; regardless we cannot say that that sum is zero.

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