AP Calculus BC

Integral of x^3 e^x: Parts Three Times

The integral of x cubed times e to the x is x cubed minus 3x squared plus 6x minus 6, all times e to the x, plus C. Integration by parts drops the power by one each pass, so a cubic needs three passes and the signs alternate.

∫x3ex dx=(x3−3x2+6x−6)ex+C\int x^{3}e^{x}\,dx = \left(x^{3}-3x^{2}+6x-6\right)e^{x} + C

Three passes, alternating signs

Each pass takes uu to be the polynomial, so its degree drops by one and the exponential is unchanged.

∫xnexdx=xnex−n∫xn−1exdx\int x^{n}e^{x}dx = x^{n}e^{x} - n\int x^{n-1}e^{x}dx

Applying it three times produces the coefficients 1,−3,6,−61, -3, 6, -6, which are 3!k!\frac{3!}{k!} with alternating signs.

The tabular shortcut

Repeated parts against an exponential is exactly the case the tabular method handles: list derivatives of x3x^{3} down one column, antiderivatives of exe^{x} down the other, and multiply diagonally with alternating signs. It gives the same four terms with far less writing.

Common mistakes

  • Losing an alternating sign, which is the dominant error on multi-pass parts.
  • Stopping after two passes and leaving an integral unevaluated.
  • Checking by differentiating only the first term.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

What is the integral of x^3 e^x?

It is (x3−3x2+6x−6)ex+C\left(x^{3}-3x^{2}+6x-6\right)e^{x} + C.

How many times do I apply parts?

Three, once per degree of the polynomial.