AP Calculus AB and BC

Integral of e^x: Answer, Proof, and Mistakes

Q: What is the integral of e^x?

\int e^x \,dx = e^x + C. The natural exponential is its own antiderivative, because \frac{d}{dx}\left[e^x\right] = e^x. The constant of integration is part of the answer for an indefinite integral.

Q: Why does e^x stay the same when you integrate it?

Because integration reverses differentiation, and e^x is unchanged by differentiation. Up to a constant multiple it is the only function with that property: the solutions of \frac{dy}{dx} = y are exactly y = Ae^x. Reversing a rule that changes nothing also changes nothing, apart from the + C.

Q: What is the integral of e^(kx)?

\int e^{kx} \,dx = \frac{1}{k}e^{kx} + C for k \neq 0. Substituting u = kx gives du = k\,dx, so the \frac{1}{k} comes out front. For example \int e^{3x} \,dx = \frac{1}{3}e^{3x} + C and \int e^{-x} \,dx = -e^{-x} + C.

Q: Do you still need + C on the integral of e^x?

Yes, on an indefinite integral. Every function of the form e^x + C has derivative e^x, so writing e^x alone names one antiderivative instead of the whole family. On a definite integral the constant cancels between the bounds, so it is not written there.

Q: What is the integral of e^(x^2)?

There is no elementary antiderivative for e^{x^2}, so no AP technique produces a closed form. Substitution works only when the inside derivative is already present, which is why \int x e^{x^2} \,dx = \frac{1}{2}e^{x^2} + C is fine while the bare e^{x^2} is not. On BC you would handle it with a Maclaurin series, or numerically.

The integral of e^x is e^x + C. The natural exponential is its own antiderivative, since the derivative of e^x is e^x, so antidifferentiating returns the same function plus a constant of integration. When the exponent is linear, divide by the coefficient: e^(kx) integrates to (1/k)e^(kx) + C.

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

Why e^x is its own antiderivative

An antiderivative of $f$ is any function $F$ with $F'(x) = f(x)$ . So the question "what integrates to $e^x$ ?" is really the question "what has derivative $e^x$ ?", and the natural exponential answers it about itself.

\frac{d}{dx}\left[e^x\right] = e^x

Read that rule backwards and you have the indefinite integral. Any two antiderivatives of the same function differ by a constant, so the general antiderivative carries $+ C$ .

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

Check it the way the AP exam expects you to check any antiderivative: differentiate your answer and see whether the integrand comes back. Here $\frac{d}{dx}\left[e^x + C\right] = e^x + 0 = e^x$ .

Only the exponential does this

Up to a constant multiple, $e^x$ is the only function equal to its own derivative: the solutions of $\frac{dy}{dx} = y$ are exactly $y = Ae^x$ . That is why base $e$ gets the clean rule while a general base picks up a logarithm, $\int a^x \,dx = \frac{a^x}{\ln a} + C$ .

The rule for e^{kx} and other linear exponents

As soon as the exponent is $kx$ rather than $x$ , the answer picks up a factor of $\frac{1}{k}$ . This is the chain rule run in reverse, and substitution makes it explicit.

\int e^{kx} \,dx = \frac{1}{k}e^{kx} + C, \quad k \neq 0

Let $u = kx$ , so $du = k\,dx$ and $dx = \frac{du}{k}$ . The integral becomes $\frac{1}{k}\int e^u \,du = \frac{1}{k}e^u + C$ , and substituting back gives the rule. The division is the tell: you divide by the derivative of the inside, never multiply.

\int e^{3x} \,dx = \frac{1}{3}e^{3x} + C

A constant added inside the exponent behaves differently. Since $e^{x+2} = e^2 \cdot e^x$ and $e^2$ is just a number, the coefficient on $x$ is still 1 and nothing gets divided.

\int e^{x+2} \,dx = e^{x+2} + C

A nonlinear exponent is a different problem

$\int e^{x^2} \,dx$ has no elementary antiderivative, so no AP technique produces a closed form for it. Substitution rescues the version that already carries the inside derivative: with $u = x^2$ and $du = 2x\,dx$ , $\int x e^{x^2} \,dx = \frac{1}{2}e^{x^2} + C$ . The stray $x$ is what makes it work.

Where the integral of e^x shows up on the AP exam

The rule is introduced in Unit 6, Topic 6.8 (Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation), as one of the basic antiderivatives you recall rather than derive. Unit 6 carries a weighting of 15 to 20 percent on both AB and BC, the largest single unit on either exam.

From there it turns up as a building block rather than as a problem in its own right.

Topic 6.9 (Integrating Using Substitution): every $e^{kx}$ and $x e^{x^2}$ integral reduces to $\int e^u \,du$ .
Topic 6.7 (The Fundamental Theorem of Calculus and Definite Integrals): $e^x$ is a favorite integrand because the antiderivative is instant and the arithmetic stays clean.
Topic 6.11 (Integrating Using Integration by Parts), a BC topic: $e^x$ is the standard choice for $dv$ precisely because integrating it changes nothing.
Topic 7.8 (Exponential Models with Differential Equations), in a unit weighted 5 to 10 percent: separating $\frac{dy}{dt} = ky$ leans on the same exponential family.

A typical definite-integral appearance evaluates in one line by the Fundamental Theorem.

\int_0^1 e^x \,dx = \left[e^x\right]_0^1 = e - 1 \approx 1.718

Common mistakes with the integral of e^x

Applying the power rule. $\int e^x \,dx$ is not $\frac{e^{x+1}}{x+1}$ . The power rule is for a variable base with a constant exponent, $x^n$ ; here the base is constant and the variable sits in the exponent, so a different rule applies.
Dropping the constant of integration. An indefinite integral names a whole family of antiderivatives, and the missing $+ C$ costs you the general solution on any differential-equation question.
Forgetting to divide for $e^{kx}$ . $\int e^{5x} \,dx = \frac{1}{5}e^{5x} + C$ , not $e^{5x} + C$ ; differentiating the wrong answer produces a stray factor of 5 that exposes it immediately.
Multiplying by the inside derivative instead of dividing. Differentiating multiplies by $u'$ , so antidifferentiating divides by it. Reversing a rule reverses the operation.
Treating $e^{x^2}$ like $e^{3x}$ . Substitution only closes when $du$ is already present up to a constant. With $u = x^2$ you need an $x$ factor, and $\int e^{x^2} \,dx$ never supplies one.
Mixing up $e^x$ with its inverse. $\int e^x \,dx = e^x + C$ , while $\int \frac{1}{x} \,dx = \ln|x| + C$ . Being inverse functions does not make them share an integration rule.

Worked examples

Example 1. A constant multiple passes straight through the integral sign.

\int 5e^x \,dx = 5e^x + C

Example 2. Sums split, and each piece uses its own basic antiderivative.

\int \left(e^x + \cos x\right) \,dx = e^x + \sin x + C

Example 3. Substitution with $u = x^2$ , $du = 2x\,dx$ , so the $x$ out front supplies $\frac{1}{2}\,du$ .

\int x e^{x^2} \,dx = \frac{1}{2}e^{x^2} + C

Example 4. A negative coefficient divides just like any other, which is where the minus sign in exponential-decay problems comes from.

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

Example 5. A definite integral whose bounds are chosen so the exponential undoes the logarithm.

\int_0^{\ln 2} e^x \,dx = \left[e^x\right]_0^{\ln 2} = 2 - 1 = 1

Verify by differentiating

Every answer above can be checked in a few seconds: differentiate it and compare with the integrand. On a free-response question that check is worth the time, because a dropped $\frac{1}{k}$ or a missing minus sign shows up instantly.

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?

$\int e^x \,dx = e^x + C$ . The natural exponential is its own antiderivative, because $\frac{d}{dx}\left[e^x\right] = e^x$ . The constant of integration is part of the answer for an indefinite integral.

Why does e^x stay the same when you integrate it?

Because integration reverses differentiation, and $e^x$ is unchanged by differentiation. Up to a constant multiple it is the only function with that property: the solutions of $\frac{dy}{dx} = y$ are exactly $y = Ae^x$ . Reversing a rule that changes nothing also changes nothing, apart from the $+ C$ .

What is the integral of e^(kx)?

$\int e^{kx} \,dx = \frac{1}{k}e^{kx} + C$ for $k \neq 0$ . Substituting $u = kx$ gives $du = k\,dx$ , so the $\frac{1}{k}$ comes out front. For example $\int e^{3x} \,dx = \frac{1}{3}e^{3x} + C$ and $\int e^{-x} \,dx = -e^{-x} + C$ .

Do you still need + C on the integral of e^x?

Yes, on an indefinite integral. Every function of the form $e^x + C$ has derivative $e^x$ , so writing $e^x$ alone names one antiderivative instead of the whole family. On a definite integral the constant cancels between the bounds, so it is not written there.

What is the integral of e^(x^2)?

There is no elementary antiderivative for $e^{x^2}$ , so no AP technique produces a closed form. Substitution works only when the inside derivative is already present, which is why $\int x e^{x^2} \,dx = \frac{1}{2}e^{x^2} + C$ is fine while the bare $e^{x^2}$ is not. On BC you would handle it with a Maclaurin series, or numerically.