Why is the integral of e^x equal to e^x itself?

Because e is uniquely defined as the base whose exponential function has a derivative equal to itself. Therefore \frac{d}{dx}(e^x) = e^x, which means the antiderivative must also be e^x.

What does the + C mean in the formula?

The constant C represents all possible vertical shifts of the antiderivative. Since the derivative of any constant is zero, adding or subtracting any constant to e^x still gives a valid antiderivative, so C accounts for this infinite family of solutions.

How do I integrate e^x with a coefficient, like \int 2e^{x}\,dx?

Factor the coefficient out first: \int 2e^{x}\,dx = 2 \int e^{x}\,dx = 2e^{x} + C. The constant multiple rule lets you pull scalars outside the integral sign, simplifying the problem to the basic case.

Is \int e^{x}\,dx different from \int e^{-x}\,dx?

Yes. For \int e^{-x}\,dx, use U-Substitution with u = -x and du = -dx to get \int e^{-x}\,dx = -e^{-x} + C. The negative sign in the exponent requires the chain rule's reverse, which substitution handles.

Integral of e^x

The integral of e^x equals e^x plus C, the essential antiderivative for solving exponential growth, decay, and differential equations.

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

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What each symbol means

What Integral of e^x takes

Symbol	Meaning
$x$	The variable with respect to which we integrate; it represents the exponent in the function $e^x$ , and if confused with a multiplicative factor you will completely misunderstand the function being integrated.
$C$	The constant of integration representing the arbitrary constant that appears in all antiderivatives; if you forget to include it, you lose the entire family of solutions and are left with just one specific function.
$e$	Euler's number (approximately 2.718), the unique base of the natural exponential function whose derivative equals itself; confusing it with the variable $x$ or treating it as just a number multiplied by $x$ will lead to incorrect integrals.

When to use it

Reach for this formula whenever you need to reverse an exponential function to find its antiderivative or solve a differential equation involving $e^x$ .

Level

Usually taught in: Calculus I

Worked examples

1. Find the antiderivative of a constant multiple

Problem

Find

\int 3e^{x}\,dx

$\int 3e^{x}\,dx = 3 \int e^{x}\,dx$
Factor the constant 3 out of the integral using the constant multiple rule.
$3 \int e^{x}\,dx = 3 e^{x}$
Apply the integral formula: $\int e^x \,dx = e^x$ .
$3 e^{x} + C$
Add the constant of integration to account for all possible antiderivatives.

Answer: $3e^{x} + C$

This problem shows that a constant multiplying $e^x$ comes straight out of the integral sign. You find the antiderivative of $e^x$ and then multiply by the constant—a direct application of linearity of integration.

2. Integrate a sum involving exponentials and polynomials

Problem

Find

\int (e^{x} + 2x)\,dx

$\int (e^{x} + 2x)\,dx = \int e^{x}\,dx + \int 2x\,dx$
Split the sum using linearity: integrate each term separately.
$\int e^{x}\,dx = e^{x}$
The integral of $e^x$ is simply $e^x$ .
$\int 2x\,dx = 2 \cdot \frac{x^2}{2} = x^2$
Integrate $2x$ using the power rule: $\int 2x \,dx = 2 \cdot \frac{x^2}{2} = x^2$ .
$e^{x} + x^2 + C$
Combine the antiderivatives and add the single constant of integration for the full answer.

Answer: $e^{x} + x^2 + C$

Linearity of integration means you can split a sum and integrate each piece separately. Here, $e^x$ integrates to itself, while the polynomial term uses the power rule. Summing with a single arbitrary constant gives the complete answer.

3. Apply exponential integration to a gaming scenario

Problem

In a battle simulator game, a super-charged opponent's attack power grows at rate

r(t) = 5e^{t}

damage per second, where

t

is seconds. If the opponent has already dealt 20 damage before charging up, what is the total damage

D(t)

after charging for

t

seconds?

$D(t) = 20 + \int 5e^{t}\,dt$
The total damage is the initial 20 damage plus the accumulation from the charging period.
$\int 5e^{t}\,dt = 5 \int e^{t}\,dt = 5 e^{t} + C$
Factor out 5 and integrate: $\int e^t \,dt = e^t$ , giving $5e^t$ plus an arbitrary constant.
$D(t) = 20 + 5 e^{t} + C$
Substitute back: the total damage after $t$ seconds is the initial damage plus the accumulated damage.
$D(0) = 20 + 5 e^{0} + C = 20 + 5 + C = 25 + C$
At $t = 0$ (before charging), the damage should be exactly 20, so $25 + C = 20$ , giving $C = -5$ .
$D(t) = 20 + 5e^{t} - 5 = 5e^{t} + 15$
Substitute $C = -5$ to get the final formula relating damage to charge time.

Answer: $D(t) = 5e^{t} + 15$

This problem uses the integral to accumulate a time-varying quantity: total damage accumulates by integrating the rate. The constant of integration is determined by an initial condition (20 damage at $t=0$ ), showing how antiderivatives solve real-world scenarios where you know rates and need totals.

Common mistakes

Where Integral of e^x usually goes wrong

The mistake	Why it is wrong	The fix
$\int e^{x}\,dx = \frac{e^{x}}{x} + C$	Some students incorrectly apply the power rule to exponentials, treating $e^x$ as if the exponent were a power and trying to divide by it.	Recognize that $e^x$ is special—the derivative of $e^x$ is $e^x$ itself, so the antiderivative is just $e^x + C$ , not divided by anything.
$\int e^{x}\,dx = e^{x}$ (without the constant $C$ )	Students forget that antiderivatives are determined only up to a constant, because the derivative of any constant is zero.	Always add the constant of integration: $\int e^{x}\,dx = e^{x} + C$ .
Confusing $\int e^{x}\,dx$ with $e \cdot \int x \,dx = e \cdot \frac{x^2}{2} + C$	Some students misread the notation and think $e$ is multiplied by $x$ instead of being the base of an exponential.	Remember that $e^x$ means $e$ raised to the power $x$ , not $e$ times $x$ . The integral is $e^x + C$ , not a quadratic.

Tips and when to use something else

This formula only works for the base $e$ . If you see a different base like $2^x$ , use the formula $\int a^{x}\,dx = \frac{a^{x}}{\ln a} + C$ instead.
Always include $+ C$ in indefinite integrals—it represents the entire family of antiderivatives that all differ by a constant.
For exponents like $\int e^{2x}\,dx$ , use U-Substitution to handle the chain rule; directly applying the formula won't work for composite functions.
When using definite integrals like $\int_a^b e^{x}\,dx$ , the constant $C$ cancels out naturally, giving you $e^{b} - e^{a}$ .

Related formulas

Indefinite IntegralThe indefinite integral finds all functions whose derivative equals a given function, expressed as an antiderivative plus an arbitrary constant.U-SubstitutionU-substitution simplifies integrals by transforming them into easier forms using a change of variables, ideal when you spot a function and its derivative.Fundamental Theorem of CalculusCompute the exact value of a definite integral using the antiderivative—connect area under a curve to the accumulation of change.Separable Differential EquationsSolve differential equations where the rate of change is a product of a function of x and a function of y by separating variables and integrating.

Frequently asked questions

Why is the integral of $e^x$ equal to $e^x$ itself?: Because $e$ is uniquely defined as the base whose exponential function has a derivative equal to itself. Therefore $\frac{d}{dx}(e^x) = e^x$ , which means the antiderivative must also be $e^x$ .
What does the $+ C$ mean in the formula?: The constant $C$ represents all possible vertical shifts of the antiderivative. Since the derivative of any constant is zero, adding or subtracting any constant to $e^x$ still gives a valid antiderivative, so $C$ accounts for this infinite family of solutions.
How do I integrate $e^x$ with a coefficient, like $\int 2e^{x}\,dx$ ?: Factor the coefficient out first: $\int 2e^{x}\,dx = 2 \int e^{x}\,dx = 2e^{x} + C$ . The constant multiple rule lets you pull scalars outside the integral sign, simplifying the problem to the basic case.
Is $\int e^{x}\,dx$ different from $\int e^{-x}\,dx$ ?: Yes. For $\int e^{-x}\,dx$ , use U-Substitution with $u = -x$ and $du = -dx$ to get $\int e^{-x}\,dx = -e^{-x} + C$ . The negative sign in the exponent requires the chain rule's reverse, which substitution handles.

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Reviewed 2026-09-18