AP Calculus AB and BC

Integral of e^(4x): Answer, Proof, and Steps

Q: Why divide by 4 and not multiply?

Because the derivative of e^{4x} multiplies by 4. To reverse a multiplication by 4, integration divides by 4, so \int e^{4x}\,dx = \frac{e^{4x}}{4} + C. Multiplying would double down on the error.

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

\int e^{kx}\,dx = \frac{e^{kx}}{k} + C for any constant k \neq 0. The \frac{1}{k} cancels the inner coefficient the chain rule produces on differentiation.

Q: Does a definite integral of e^(4x) use the same antiderivative?

Yes. For example \int_0^1 e^{4x}\,dx = \left[\frac{e^{4x}}{4}\right]_0^1 = \frac{e^4 - 1}{4}. Find the antiderivative first, then evaluate at the limits.

The integral of e^(4x) is (e^(4x))/4 + C. Because differentiating e^(4x) multiplies by the inner coefficient 4, integrating divides by that 4 to cancel it. You can confirm the result by differentiating (e^(4x))/4 and recovering e^(4x).

\int e^{4x}\,dx = \frac{e^{4x}}{4} + C

Undo the chain rule factor

Differentiating $e^{4x}$ brings the inner coefficient 4 out front, by the chain rule. Integration reverses that, so the antiderivative carries a dividing 4 to cancel the factor differentiation would create.

\frac{d}{dx}e^{4x} = 4e^{4x}

\int e^{4x}\,dx = \frac{e^{4x}}{4} + C

The pattern $\int e^{kx}\,dx = \frac{e^{kx}}{k} + C$ holds for any nonzero constant $k$ .

The same result by substitution

Substitution makes the division explicit. Let $u = 4x$ , so $du = 4\,dx$ and $dx = \frac{du}{4}$ .

\int e^{4x}\,dx = \int e^{u}\cdot\frac{du}{4} = \frac{1}{4}\int e^{u}\,du = \frac{1}{4}e^{u} + C = \frac{e^{4x}}{4} + C

Differentiating $\frac{e^{4x}}{4}$ gives $\frac{1}{4}\cdot 4e^{4x} = e^{4x}$ , confirming the antiderivative.

The mistake students make

The most common answer is $e^{4x}$ with no division, treating $e^{4x}$ like $e^x$ , which integrates to itself. Differentiating $e^{4x}$ shows why that fails: it returns $4e^{4x}$ , four times too large. Others multiply by 4 instead of dividing, writing $4e^{4x}$ , which differentiates to $16e^{4x}$ .

Divide by the inner coefficient

For $\int e^{kx}\,dx$ divide by $k$ ; for $\int \sin kx\,dx$ or $\int\cos kx\,dx$ the same $\frac{1}{k}$ appears. It always cancels the coefficient the chain rule would introduce.

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

Why divide by 4 and not multiply?

Because the derivative of $e^{4x}$ multiplies by 4. To reverse a multiplication by 4, integration divides by 4, so $\int e^{4x}\,dx = \frac{e^{4x}}{4} + C$ . Multiplying would double down on the error.

What is the integral of e^(kx) in general?

$\int e^{kx}\,dx = \frac{e^{kx}}{k} + C$ for any constant $k \neq 0$ . The $\frac{1}{k}$ cancels the inner coefficient the chain rule produces on differentiation.

Does a definite integral of e^(4x) use the same antiderivative?

Yes. For example $\int_0^1 e^{4x}\,dx = \left[\frac{e^{4x}}{4}\right]_0^1 = \frac{e^4 - 1}{4}$ . Find the antiderivative first, then evaluate at the limits.