AP Calculus AB and BC

Integral of e^(-x): Answer, Proof, and Steps

Q: Why is the integral of e^(-x) negative?

Because the exponent -x has slope -1, and integrating divides by that slope. The factor \frac{1}{-1} = -1 becomes the leading minus in -e^{-x} + C.

Q: How does the integral of e^(-x) compare to its derivative?

They match: the derivative of e^{-x} is -e^{-x}, and the antiderivative is -e^{-x} + C. The inner slope of -1 makes the function reproduce itself with a sign each time.

The integral of e^(-x) is -e^(-x) + C. The exponent -x has slope -1, so you divide by that inner coefficient, which makes the sign negative. Differentiating -e^(-x) gives -e^(-x) times -1 = e^(-x), which confirms the answer.

\int e^{-x}\,dx = -e^{-x} + C

Dividing by the inner coefficient

The integrand is a composite: the natural exponential wrapped around $-x$ . A u-substitution on the inside function handles it.

u = -x \quad\Longrightarrow\quad \frac{du}{dx} = -1 \quad\Longrightarrow\quad dx = -du

Replace the exponent and the differential, then pull the constant outside the integral.

\int e^{-x}\,dx = \int e^{u}\cdot(-du) = -\int e^{u}\,du = -e^{u} + C

Substituting $u = -x$ back gives the answer in the original variable.

\int e^{-x}\,dx = -e^{-x} + C

The general pattern

$\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$ for any nonzero $k$ . With $k = -1$ the reciprocal $\frac{1}{-1}$ is the leading minus.

Why the sign flips

Differentiating the answer shows where the minus earns its place. The chain rule attaches a factor of $-1$ to the derivative of $e^{-x}$ .

\frac{d}{dx}\left(e^{-x}\right) = e^{-x}\cdot\frac{d}{dx}(-x) = -e^{-x}

So $e^{-x}$ by itself differentiates to the negative of itself. The leading minus on the antiderivative cancels that stray sign.

\frac{d}{dx}\left(-e^{-x}\right) = -\left(-e^{-x}\right) = e^{-x}

Common mistakes

Forgetting the negative and writing $e^{-x} + C$ , which differentiates to $-e^{-x}$ , the wrong sign.
Applying the power rule. The variable is in the exponent, not the base, so this is not $\frac{e^{-x+1}}{-x+1}$ .
Dividing by $x$ instead of the coefficient $-1$ . Only the constant slope of the exponent, here $-1$ , sets the factor.

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 e^(-x) from 0 to infinity?

$\int_0^{\infty} e^{-x}\,dx = \left[-e^{-x}\right]_0^{\infty} = 0 - (-1) = 1$ . As $x \to \infty$ , $e^{-x} \to 0$ , so the improper integral converges to $1$ .

Why is the integral of e^(-x) negative?

Because the exponent $-x$ has slope $-1$ , and integrating divides by that slope. The factor $\frac{1}{-1} = -1$ becomes the leading minus in $-e^{-x} + C$ .

How does the integral of e^(-x) compare to its derivative?

They match: the derivative of $e^{-x}$ is $-e^{-x}$ , and the antiderivative is $-e^{-x} + C$ . The inner slope of $-1$ makes the function reproduce itself with a sign each time.