# Integral of eˣ: Formula, Steps and Examples

> The integral of eˣ is eˣ + C. See why eˣ is its own antiderivative, a graph of the area, definite and improper integrals and worked examples.

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

$$ \int e^x\,dx = e^x + C $$

## At a glance

- **Antiderivative:** $e^x + C$
- **From 0 to 1:** $e - 1 \approx 1.718$
- **From −∞ to 0:** $1$
- **With a linear inside:** $\int e^{kx+b}\,dx = \frac{1}{k}e^{kx+b} + C$
- **Derivative:** $\frac{d}{dx}e^x = e^x$

## Key takeaways

- The integral of $e^x$ is $e^x + C$, because $e^x$ is its own derivative.
- The area under $e^x$ from $a$ to $b$ is simply $e^b - e^a$.
- With a linear inside, divide by the coefficient of $x$: $\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$.
- Do not use the power rule on $e^x$. The variable is in the exponent, so $\frac{e^{x+1}}{x+1}$ is wrong.

The **integral of $e^x$ is $e^x + C$**. The natural exponential is the only function (apart from multiples of it) that is its own derivative, so it is also its own antiderivative.

$$\int e^x\,dx = e^x + C$$

## Graph of eˣ with the area from 0 to 1

```graph
title: 'y = eˣ with the area from 0 to 1 shaded'
caption: 'The shaded area is $e^1 - e^0 = e - 1 \approx 1.718$. Because the antiderivative is $e^x$ itself, the area up to any point $b$ (starting from $-\infty$) equals the height $e^b$.'
x: [-3, 2]
y: [-0.5, 6]
functions:
  - expr: e^x
    label: y = eˣ
shade:
  expr: e^x
  from: 0
  to: 1
points:
  - [0, 1, '(0, 1)']
  - [1, 2.718, '(1, e)']
```

## Why the integral of eˣ is eˣ

An antiderivative of $f$ is any function whose derivative is $f$. The [derivative of eˣ](/derivatives/e-to-the-x) is $e^x$, so $e^x$ works:

$$\frac{d}{dx}\left(e^x + C\right) = e^x$$

That is the whole proof. The $+ C$ is there because any constant has derivative zero, so $e^x + 5$ and $e^x - 2$ are antiderivatives too. The [antiderivatives guide](/blog/antiderivatives) explains why every antiderivative has this form.

## Step by step: integrals built from eˣ

1. **Pull out constants.** $\int 4e^x\,dx = 4e^x + C$, by the [constant multiple rule](/rules/constant-multiple-rule).
2. **Split sums.** $\int (e^x + x^2)\,dx = e^x + \frac{x^3}{3} + C$.
3. **Rewrite shifted exponents.** $e^{x+2} = e^2\cdot e^x$, so $\int e^{x+2}\,dx = e^{x+2} + C$.
4. **Divide by the inside slope.** If the exponent is $kx + b$, the answer is $\frac{1}{k}e^{kx+b} + C$. The [integral of e²ˣ](/integrals/e-to-the-2x) page covers this case in detail.

## Check by differentiating

For $\frac{1}{k}e^{kx+b}$, the [chain rule](/rules/chain-rule) gives $\frac{1}{k}\cdot e^{kx+b}\cdot k = e^{kx+b}$. The factor $\frac{1}{k}$ is exactly what cancels the $k$ from the chain rule.

## Definite and improper integrals

By the [Fundamental Theorem of Calculus](/theorems/fundamental-theorem-of-calculus-part-2), $\int_a^b e^x\,dx = e^b - e^a$.

| Integral | Value |
| --- | --- |
| $\int_0^1 e^x\,dx$ | $e - 1 \approx 1.718$ |
| $\int_0^{\ln 2} e^x\,dx$ | $2 - 1 = 1$ |
| $\int_{-1}^{1} e^x\,dx$ | $e - \frac{1}{e} \approx 2.350$ |
| $\int_{-\infty}^{0} e^x\,dx$ | $1 - \lim_{a\to-\infty}e^a = 1$ |

The last one is an [improper integral](/blog/improper-integrals). The left tail of $e^x$ goes on forever, but it shrinks so fast that its total area is exactly $1$.

## Worked examples

**Example 1: a sum.** $\displaystyle\int (3e^x - 2x)\,dx = 3e^x - x^2 + C$.

**Example 2: e in the base and the exponent.** $\displaystyle\int e^{x}e^{2}\,dx = e^2 e^x + C = e^{x+2} + C$. Combine exponents only when the bases match.

**Example 3: a substitution.** $\displaystyle\int \frac{e^x}{1 + e^x}\,dx$. Let $u = 1 + e^x$, $du = e^x\,dx$:

$$\int\frac{du}{u} = \ln(1 + e^x) + C$$

No absolute value is needed since $1 + e^x > 0$.

**Example 4: the exponent is a function.** $\displaystyle\int 2x\,e^{x^2}\,dx$. With $u = x^2$, $du = 2x\,dx$, this is $\int e^u\,du = e^{x^2} + C$. This only works because the $2x$ is there. Without it, $\int e^{x^2}\,dx$ has no elementary answer; see the [integral of e^(x²)](/integrals/e-to-the-x-squared).

**Example 5: average value.** The average of $e^x$ on $[0, 1]$ is $\frac{1}{1-0}\int_0^1 e^x\,dx = e - 1 \approx 1.718$.

## Common mistakes

1. **Using the power rule.** $\int e^x\,dx \ne \frac{e^{x+1}}{x+1}$. The power rule is for $x^n$, where the variable is the base.
2. **Forgetting to divide by $k$.** $\int e^{5x}\,dx = \frac{1}{5}e^{5x} + C$, not $e^{5x} + C$.
3. **Dividing by a non-constant.** $\int e^{x^2}\,dx \ne \frac{e^{x^2}}{2x}$. Differentiate that and you will not get $e^{x^2}$ back.
4. **Dropping $+ C$** on an indefinite integral.

For other bases such as $2^x$, see the [integral of 2ˣ](/integrals/2-to-the-x). The [natural exponential function](/functions/natural-exponential-function) page explains where $e$ comes from.

## Frequently asked questions

### What is the integral of eˣ?

It is $e^x + C$.

### What is the integral of eˣ from 0 to 1?

It is $e^1 - e^0 = e - 1 \approx 1.718$.

### What is the integral of e^(x+3)?

It is $e^{x+3} + C$. The inside $x + 3$ has derivative $1$, so nothing changes. You can also write $e^{x+3} = e^3 e^x$ and treat $e^3$ as a constant.

### Why is the integral of eˣ equal to eˣ?

Because the derivative of $e^x$ is $e^x$. Integration undoes differentiation, so a function that is its own derivative is also its own antiderivative, up to the constant $C$.