Evaluate the integral # int \ xe^(-ax^2) \ dx #?

Question

12663 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-25T19:30:46

# int \ xe^(-ax^2) \ dx = -1/(2a) \ e^(-ax^2) +C #

I assume we seek:

# I = int \ xe^(-ax^2) \ dx #

We can perform a substitution:

# u = -ax^2 => (du)/(dx) = -2ax #

Then we can rewrite the integral and perform a substitution

# I = -1/(2a) \ int \ (-2ax)e^(-ax^2) \ dx #
# \ \ = -1/(2a) \ int \ e^u \ du #
# \ \ = -1/(2a) \ e^u +C #

And restoring the substitution we have:

# I = -1/(2a) \ e^(-ax^2) +C #