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

> The integral of e^(−x) is −e^(−x) + C. See the substitution, why the area from 0 to infinity is 1, a graph and examples like x e^(−x).

Source: https://practicecalculus.com/integrals/e-to-the-negative-x
Updated: 2026-11-06

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

## At a glance

- **Antiderivative:** $-e^{-x} + C$
- **From 0 to ∞:** $1$
- **From 0 to 1:** $1 - \frac{1}{e} \approx 0.632$
- **General form:** $\int e^{-kx}\,dx = -\frac{1}{k}e^{-kx} + C$
- **Method:** Substitution $u = -x$, $du = -dx$

## Key takeaways

- The integral of $e^{-x}$ is $-e^{-x} + C$. The minus sign comes from the inside derivative $-1$.
- The total area under $e^{-x}$ from $0$ to $\infty$ is exactly $1$.
- In general, $\int_0^\infty e^{-kx}\,dx = \frac{1}{k}$ for $k > 0$, which is why $ke^{-kx}$ is a probability density.
- $\int x e^{-x}\,dx = -(x+1)e^{-x} + C$, by integration by parts.

The **integral of $e^{-x}$ is $-e^{-x} + C$**. The only change from the [integral of eˣ](/integrals/e-to-the-x) is the minus sign, which comes from the $-1$ in the exponent.

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

## Graph of e⁻ˣ and its area

```graph
title: 'y = e^(−x) with the area from 0 to 4 shaded'
caption: 'The shaded area is $1 - e^{-4} \approx 0.982$. If you keep going to infinity, the area approaches exactly $1$.'
x: [-1, 6]
y: [-0.5, 2.5]
functions:
  - expr: e^(-x)
    label: y = e^(−x)
shade:
  expr: e^(-x)
  from: 0
  to: 4
points:
  - [0, 1, '(0, 1)']
```

$e^{-x}$ 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](/integrals/e-to-the-2x) page. For decay rates $k > 0$:

$$\int e^{-kx}\,dx = -\frac{1}{k}e^{-kx} + C$$

## Check by differentiating

$$\frac{d}{dx}\left(-e^{-x}\right) = -e^{-x}\cdot(-1) = e^{-x}$$

The two minus signs cancel. See the [derivative of e⁻ˣ](/derivatives/e-to-the-negative-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](/blog/improper-integrals), so take a limit:

$$\int_0^\infty e^{-x}\,dx = \lim_{b\to\infty}\left(1 - e^{-b}\right) = 1$$

**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](/rules/integration-by-parts-rule) with $u = x$ and $dv = e^{-x}\,dx$, so $du = dx$ and $v = -e^{-x}$:

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

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

$$\int_0^4 40e^{-0.5t}\,dt = -80\left(e^{-2} - 1\right) = 80\left(1 - e^{-2}\right) \approx 69.2 \text{ mg}$$

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

$$\int\cosh x\,dx = \frac{e^x - e^{-x}}{2} + C = \sinh x + C$$

## 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)](/integrals/e-to-the-x-squared).

For more on products like $xe^{-x}$, see the [integral of x eˣ](/integrals/x-e-to-the-x).

## Frequently asked questions

### What is the integral of e^(−x)?

It is $-e^{-x} + C$.

### What is the integral of e^(−x) from 0 to infinity?

It is $1$. The antiderivative $-e^{-x}$ goes from $-1$ at $x = 0$ to $0$ as $x \to \infty$, so the integral is $0 - (-1) = 1$.

### What is the integral of x e^(−x)?

It is $-(x+1)e^{-x} + C$. Use parts with $u = x$ and $dv = e^{-x}\,dx$. From $0$ to $\infty$ it equals $1$.

### Why is there a minus sign in the integral of e^(−x)?

Because the derivative of $-e^{-x}$ is $-e^{-x}\cdot(-1) = e^{-x}$. Without the minus sign, the chain rule would leave you with $-e^{-x}$.