MML Discourse archived in May, 2026

A u-substitution integral

mark

Compute

\int_0^b x e^{-x^2}\, dx.

Be sure, to

  • use u-substitution with proper notation,
  • change your bounds of integration, and
  • express your answer in terms of the unspecified parameter b.
User 026

Using u-substitution to solve the function:

\int_{0}^{b}xe^{-x^2}

I chose u to be u = -x^2 which makes du = -2x \, dx and -\frac{du}{2} = x \, dx.

To change our bounds we need to know what x is at our lower and upper bounds. When x is 0 our it is 0 and when x is equal to b x is -b^2 so our bounds become \int_{0}^{-b^2}. We can then write our integral in terms of u with our adjusted bounds.

-\frac{1}{2} \int_{0}^{-b^2} e^{u} \, du

Next subtract our function at it's lower bound from our function at it's upper bound. e^{0^2} Is one so our integral becomes:

-\frac{1}{2} \left[ e^{u} \right]_{0}^{-b^2} = -\frac{1}{2} \left( e^{-b^2} - 1 \right)

We can factor the negative from our -\frac{1}{2} into the parenthesis to clean up our expression. (same answer I just prefer positive coefficients)

\frac{1}{2} \left( 1 - e^{-b^2} \right)
User 009
\int_0^b xe^{-x^2} dx\\ \\ u = -x^2\\ du = -2x dx\\ \frac{du}{-2x} = dx\\ \\ \frac{-1}{2}\int_0^{-b^2} xe^{u} du\\ \\ \frac{-1}{2}e^u\Bigg|_0^{-b^2}\\ \\ -1/2(e^{-b^2}-e^0)\\ \\ e^0 = 1\\ \\ -1/2(e^{-b^2}-1)\\