Evaluate the integral? : # int x^3e^(x^2) dx #
2 Answers
The answer is
Explanation:
We need the integration by parts
We perform the substitution
Let
Therefore,
We apply the integration by parts
Therefore,
We seek:
# int \ x^3e^(x^2) \ dx = 1/2(x^2-1)e^(x^2) + C #
Explanation:
We seek:
# I = int \ x^3e^(x^2) \ dx #
Note as a helper that:
# d/dx ( e^(x^2) ) = 2xe^(x^2) iff int \ 2xe^(x^2) \ dx = e^(x^2) #
So we can write the integral as:
# 2I = int \ (x^2)(2xe^(x^2)) \ dx #
We can now use the formula for Integration By Parts (IBP):
# int \ u(dv)/dx \ dx = uv - int \ v(du)/dx \ dx # , or less formally
# " " int \ u \ dv=uv-int \ v \ du #
I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation.
Let
# { (u,=x^2, => (du)/dx,=2x), ((dv)/dx,=2xe^(x^2), => v,=e^(x^2) ) :}#
Then plugging into the IBP formula:
# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #
gives us
# int \ (x^2)(2xe^(x^2)) \ dx = (x^2)(e^(x^2)) - int \ (e^(x^2))(2x) \ dx #
# :. 2I = x^2e^(x^2) - int \ 2xe^(x^2) \ dx + A #
# " " = x^2e^(x^2) - e^(x^2) + A #
# " " = (x^2-1)e^(x^2) + A #
Hence
# I = 1/2(x^2-1)e^(x^2) + C #