AP Calculus BC

Integral of x e^-x: Parts and Three Minus Signs

Q: What is the integral of x e^-x?

It is -\left(x+1\right)e^{-x} + C. Differentiating that product returns xe^{-x}, which is the fastest way to confirm the signs.

Q: Why is u = x and not u = e^-x?

Because x differentiates down to 1 and clears the polynomial factor in one step. Taking u = e^{-x} raises the power of x instead and makes the remaining integral worse.

Q: What is the integral of x e^-x from 0 to infinity?

It is 1. The antiderivative -\left(x+1\right)e^{-x} tends to 0 as x\to\infty and equals -1 at x = 0.

The integral of x times e to the negative x is minus the quantity x plus one, times e to the negative x, plus C. Integration by parts with u equal to x is what does it, and the three minus signs that appear along the way are where nearly every wrong answer comes from.

\int xe^{-x}\,dx = -\left(x+1\right)e^{-x} + C

Parts with u = x

The polynomial factor is the one that gets simpler when differentiated, so it becomes $u$ . Take $u = x$ and $dv = e^{-x}dx$ , which gives $du = dx$ and $v = -e^{-x}$ . The minus sign in $v$ is the first of the three.

\int xe^{-x}\,dx = -xe^{-x} - \int\left(-e^{-x}\right)dx = -xe^{-x} - e^{-x} + C

The leftover integral $\int e^{-x}dx$ is itself $-e^{-x}$ , which is the third minus sign. Factoring $-e^{-x}$ out of both terms gives the form worth remembering.

\int xe^{-x}\,dx = -\left(x+1\right)e^{-x} + C

The improper integral underneath

This integrand is the reason $\int_{0}^{\infty} xe^{-x}\,dx$ shows up so often. The antiderivative is bounded at infinity because $e^{-x}$ shrinks faster than $x+1$ grows, so the limit exists.

\int_{0}^{\infty} xe^{-x}\,dx = \lim_{b\to\infty}\left[-\left(x+1\right)e^{-x}\right]_{0}^{b} = 0 - \left(-1\right) = 1

Why the limit is zero

As x grows, (x+1)e^(-x) is a linear function fighting an exponential decay, and the exponential always wins. That single fact is what makes the area under x e^(-x) from 0 to infinity finite, and it equals exactly 1.

The mistakes students make

The structure of this problem is rarely the issue. The signs are.

Using $v = e^{-x}$ instead of $v = -e^{-x}$ . That produces $xe^{-x} + e^{-x} + C$ , which is $\left(x+1\right)e^{-x}$ : the correct shape with every sign reversed.
Losing the minus sign inside the second integral and answering $-xe^{-x} + e^{-x} + C$ . Differentiating that gives $xe^{-x} - 2e^{-x}$ , not the integrand.
Choosing $u = e^{-x}$ and $dv = x\,dx$ , which leaves $\int\frac{x^{2}}{2}e^{-x}dx$ , a harder integral than the original.

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 e^-x?

It is $-\left(x+1\right)e^{-x} + C$ . Differentiating that product returns $xe^{-x}$ , which is the fastest way to confirm the signs.

Why is u = x and not u = e^-x?

Because $x$ differentiates down to $1$ and clears the polynomial factor in one step. Taking $u = e^{-x}$ raises the power of $x$ instead and makes the remaining integral worse.

What is the integral of x e^-x from 0 to infinity?

It is $1$ . The antiderivative $-\left(x+1\right)e^{-x}$ tends to $0$ as $x\to\infty$ and equals $-1$ at $x = 0$ .