AP Calculus AB and BC

Integral of e^(2x): Answer, Steps, and Mistakes

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

It is \frac{1}{2}e^{2x} + C. Substituting u = 2x turns dx into \frac{du}{2}, so a factor of \frac{1}{2} comes out front of \int e^{u}\,du = e^{u} + C.

Q: Why is there a 1/2 in the answer?

Because differentiating e^{2x} gives 2e^{2x} by the chain rule. The \frac{1}{2} cancels that extra factor of 2, so that \frac{d}{dx}\left(\frac{1}{2}e^{2x}\right) = e^{2x} exactly.

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

For any nonzero constant k, \int e^{kx}\,dx = \frac{1}{k}e^{kx} + C. So \int e^{5x}\,dx = \frac{1}{5}e^{5x} + C and \int e^{-x}\,dx = -e^{-x} + C.

Q: Can I integrate e^(x^2) the same way?

No. Substituting u = x^{2} requires a factor of x in the integrand to match du = 2x\,dx, and e^{x^{2}} has none. That integral has no elementary antiderivative, so it never appears as a symbolic AP problem.

The integral of e^(2x) is (1/2)e^(2x) + C. Substituting u = 2x gives du = 2 dx, so dx = du/2 and a factor of 1/2 comes out front. Checking backward, the derivative of (1/2)e^(2x) is (1/2) times 2 times e^(2x), which is e^(2x).

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

The substitution, step by step

The integrand is a composite: the natural exponential wrapped around $2x$ . That is the signal for u-substitution, with the inside function as $u$ .

u = 2x \quad\Longrightarrow\quad \frac{du}{dx} = 2 \quad\Longrightarrow\quad dx = \frac{du}{2}

Replace both the exponent and the differential, then pull the constant $\frac{1}{2}$ outside the integral.

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

Substituting $u = 2x$ back in finishes the problem. Always return to the original variable in an indefinite integral.

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

The general pattern is worth memorizing outright: $\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$ for any nonzero constant $k$ . Every linear exponent divides by its own coefficient.

Why the 1/2 has to be there

Differentiating the answer is the fastest check, and it shows exactly where the $\frac{1}{2}$ earns its place. By the chain rule, differentiating $e^{2x}$ produces a stray factor of $2$ .

\frac{d}{dx}\left(e^{2x}\right) = e^{2x}\cdot\frac{d}{dx}(2x) = 2e^{2x}

So $e^{2x}$ alone is not an antiderivative of $e^{2x}$ ; it differentiates to twice too much. The $\frac{1}{2}$ out front is the constant that cancels that factor.

\frac{d}{dx}\left(\frac{1}{2}e^{2x} + C\right) = \frac{1}{2}\cdot 2e^{2x} = e^{2x}

This is the structural difference between $e^{x}$ and $e^{2x}$ . The base case $\int e^{x}\,dx = e^{x} + C$ needs no coefficient because the inside derivative is $1$ . As soon as the exponent has a slope other than $1$ , u-substitution introduces a reciprocal.

Integration undoes the chain rule, so it has to undo the chain rule factor too. If differentiating your answer does not reproduce the integrand exactly, the coefficient is wrong.

Where this shows up on the AP exam

The rule lives in CED Unit 6 (Integration and Accumulation of Change): Topic 6.8 covers antiderivatives from basic rules, and Topic 6.9 is integration using substitution, where $e^{kx}$ is the standard first example.

Definite integrals such as $\int_{0}^{1} e^{2x}\,dx = \frac{1}{2}\left(e^{2} - 1\right)$ , where the $\frac{1}{2}$ must survive the evaluation.
Separable differential equations in Unit 7, where integrating $e^{kt}$ produces the growth and decay solutions $y = Ce^{kt}$ .
Area and volume setups in Unit 8, where an exponential curve is the boundary of the region.
Accumulation-function problems, where $F(x) = \int_{a}^{x} e^{2t}\,dt$ is differentiated back with the Fundamental Theorem of Calculus.

On a definite integral you may either back-substitute to $x$ and use the original limits, or change the limits along with the variable. Changing limits means $x = 0$ and $x = 1$ become $u = 0$ and $u = 2$ .

\int_{0}^{1} e^{2x}\,dx = \frac{1}{2}\int_{0}^{2} e^{u}\,du = \frac{1}{2}\left(e^{2} - e^{0}\right) = \frac{e^{2} - 1}{2}

Common mistakes

Nearly every lost point on this integral is one of four errors, and each one has a one-line diagnosis.

Forgetting the $\frac{1}{2}$ and writing $e^{2x} + C$ . Differentiate to check: that gives $2e^{2x}$ , which is twice the integrand.
Multiplying by $2$ instead of dividing. Differentiation multiplies by the inside derivative, so integration divides by it.
Treating the exponent like a power rule and writing $\frac{e^{2x+1}}{2x+1}$ . The power rule applies to a variable base, never to a variable exponent.
Dropping $+ C$ on an indefinite integral, which costs the answer point even when the antiderivative itself is right.

A related trap is the look-alike $\int e^{x^{2}}\,dx$ . Substituting $u = x^{2}$ gives $du = 2x\,dx$ , and there is no spare $x$ in the integrand to absorb it, so the substitution stalls. That integral has no elementary antiderivative and is out of scope for AP Calculus.

Only a linear exponent yields a clean constant multiple. If the exponent is $x^{2}$ or any other nonlinear expression with no matching factor present, substitution fails and the problem is asking for something else, usually a series or a numerical estimate.

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^(2x)?

It is $\frac{1}{2}e^{2x} + C$ . Substituting $u = 2x$ turns $dx$ into $\frac{du}{2}$ , so a factor of $\frac{1}{2}$ comes out front of $\int e^{u}\,du = e^{u} + C$ .

Why is there a 1/2 in the answer?

Because differentiating $e^{2x}$ gives $2e^{2x}$ by the chain rule. The $\frac{1}{2}$ cancels that extra factor of $2$ , so that $\frac{d}{dx}\left(\frac{1}{2}e^{2x}\right) = e^{2x}$ exactly.

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

For any nonzero constant $k$ , $\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$ . So $\int e^{5x}\,dx = \frac{1}{5}e^{5x} + C$ and $\int e^{-x}\,dx = -e^{-x} + C$ .

Can I integrate e^(x^2) the same way?

No. Substituting $u = x^{2}$ requires a factor of $x$ in the integrand to match $du = 2x\,dx$ , and $e^{x^{2}}$ has none. That integral has no elementary antiderivative, so it never appears as a symbolic AP problem.