Integral of e^(−x): Formula, Steps and Examples

The integral of e^{-x} is -e^{-x} + C. The only change from the integral of eˣ is the minus sign, which comes from the -1 in the exponent.

Graph of e⁻ˣ and its area

e^{-x} is exponential decay: it starts at 1 when x = 0 and loses the same fraction of its height over every unit step. That steady shrinking is what makes the infinite tail add up to a finite area.

How to integrate e⁻ˣ step by step

1. Let u = -x. Then du = -dx, so dx = -du.
2. Rewrite: \displaystyle\int e^{-x}\,dx = -\int e^u\,du.
3. Integrate: -e^u + C.
4. Substitute back: -e^{-x} + C.

This is the case k = -1 of the rule \int e^{kx}\,dx = \frac{1}{k}e^{kx} + C from the integral of e²ˣ page. For decay rates k > 0:

Check by differentiating

The two minus signs cancel. See the derivative of e⁻ˣ.

Definite and improper integrals

From 0 to 1: \left[-e^{-x}\right]_0^1 = -e^{-1} + 1 = 1 - \frac{1}{e} \approx 0.632.

From 0 to ∞: this is an improper integral, so take a limit:

A decay rate k. \int_0^\infty e^{-kx}\,dx = \frac{1}{k} for k > 0. Multiplying by k gives \int_0^\infty ke^{-kx}\,dx = 1, the total probability of the exponential distribution used for waiting times.

Half-life check. \int_0^{\ln 2} e^{-x}\,dx = 1 - \frac{1}{2} = \frac{1}{2}. Half of the total area sits before x = \ln 2 \approx 0.693.

Worked examples

Example 1: a scaled exponent. \displaystyle\int e^{-x/3}\,dx = -3e^{-x/3} + C.

Example 2: x times e⁻ˣ. Use integration by parts with u = x and dv = e^{-x}\,dx, so du = dx and v = -e^{-x}:

From 0 to \infty this equals 0 - (-1) = 1, because xe^{-x} \to 0 as x \to \infty.

Example 3: drug in the bloodstream. A drug leaves the body at a rate of 40e^{-0.5t} mg per hour. The amount removed in the first 4 hours is

Example 4: hyperbolic functions. \cosh x = \frac{e^x + e^{-x}}{2}, so

Common mistakes

1. Dropping the minus sign. \int e^{-x}\,dx is not e^{-x} + C. Differentiate your answer: you should get +e^{-x}.
2. Writing \frac{e^{-x+1}}{-x+1}. That is the power rule used on an exponential, which is never valid.
3. Treating the improper integral as divergent. The interval is infinite, but the area is 1. Use a limit, do not stop at "infinity".
4. Confusing e^{-x} with e^{-x^2}. The bell curve e^{-x^2} has no elementary antiderivative; see the integral of e^(x²).

For more on products like xe^{-x}, see the integral of x eˣ.