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How do you integrate #int x^5e^(-x^3)#?

Answer 1

Samantha Adkison

#int x^5e^(-x^3)dx = -(e^(-x^3)(x^3+1))/3 +C#

Substitute #t=x^3# so that #dt=3x^2dx# and note that:

#x^5dx = x^3/3*3x^2dx = (tdt)/3#

We then have:

#int x^5e^(-x^3)dx = 1/3 int te^(-t)dt#

Now integrate by parts using #t# as integer factor:

#int te^(-t)dt = -int td(e^(-t)) =-te^(-t) + int e^(-t)dt = -e^(-t)(t+1)+C#

and undoing the substitution:

#int x^5e^(-x^3)dx = -(e^(-x^3)(x^3+1))/3 +C#

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Answer 2

Isaiah Allen

To integrate ( \int x^5e^{-x^3} ), you can use substitution method. Let ( u = -x^3 ), then ( du = -3x^2dx ). Rearranging, ( dx = \frac{-1}{3x^2}du ). Substituting these into the integral gives ( \int x^5e^{-x^3}dx = \int x^2e^u \frac{-1}{3x^2}du ). Simplifying, ( \int x^5e^{-x^3}dx = \int \frac{-1}{3} e^u du ). Now, integrate ( \int \frac{-1}{3} e^u du ) with respect to ( u ) to get ( \frac{-1}{3}e^u + C ). Substitute back ( u = -x^3 ) to get the final result: ( \frac{-1}{3}e^{-x^3} + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.