Integration Of E 2x 2: The Shortcut Worth Teaching Well
The integral of $$e^{2x}$$ is $$\tfrac{1}{2}e^{2x} + C$$, because the chain rule in reverse requires dividing by the derivative of the exponent, which is 2. In plain terms, the antiderivative of the exponential function $$e^{2x}$$ is the same expression scaled by one-half.
Core result
The standard antiderivative is $$\int e^{2x}\,dx = \tfrac{1}{2}e^{2x} + C$$. This follows from the general rule $$\int e^{ax}\,dx = \tfrac{1}{a}e^{ax} + C$$ for any nonzero constant $$a$$, which is the key pattern students should recognize rather than memorize in isolation. The result is consistent with published calculus explanations that show the same final form for $$e^{2x}$$.
Why it works
The fastest way to see the answer is to differentiate $$\tfrac{1}{2}e^{2x}$$: the derivative of $$e^{2x}$$ brings down a factor of 2, and the $$\tfrac{1}{2}$$ cancels it. That cancellation is why the chain rule and integration are inverse operations here. This is the same logic used in substitution-based solutions for $$e^{2x}$$.
- Start with $$\int e^{2x}\,dx$$.
- Let $$u = 2x$$, so $$du = 2\,dx$$ and $$dx = \tfrac{1}{2}du$$.
- Rewrite the integral as $$\tfrac{1}{2}\int e^u\,du$$.
- Integrate to get $$\tfrac{1}{2}e^u + C$$.
- Substitute back $$u = 2x$$, giving $$\tfrac{1}{2}e^{2x} + C$$.
Common mistake
A frequent error is to write $$e^{2x} + C$$ without the $$\tfrac{1}{2}$$. That misses the scaling effect created by the inner derivative, so the derivative of the proposed answer would be $$2e^{2x}$$, not $$e^{2x}$$. Another common mistake is applying the power rule, which does not apply to exponentials.
| Expression | Antiderivative | Reason |
|---|---|---|
| $$\int e^{x}\,dx$$ | $$e^{x}+C$$ | The inner derivative is 1. |
| $$\int e^{2x}\,dx$$ | $$\tfrac{1}{2}e^{2x}+C$$ | Divide by the inner derivative 2. |
| $$\int e^{ax}\,dx$$ | $$\tfrac{1}{a}e^{ax}+C$$ | General constant multiple rule. |
Quick check
Differentiate the answer to verify it: $$\frac{d}{dx}\left(\tfrac{1}{2}e^{2x}\right)=\tfrac{1}{2}\cdot 2e^{2x}=e^{2x}$$. This confirmation is useful in class, on exams, and in any setting where students need to show that their antiderivative is correct. The same verification principle is a reliable habit in calculus problem-solving.
"Integrate the outside function, then divide by the derivative of the inside."
Study takeaway
For this type of problem, the best habit is to look for an exponential with a linear inner term and then apply the reverse chain rule. Once that pattern is clear, problems like $$\int e^{5x}\,dx$$, $$\int e^{-3x}\,dx$$, and $$\int e^{ax}\,dx$$ all become immediate. The rule is compact, but the reasoning behind it is what makes it durable knowledge.
Expert answers to Integration Of E 2x 2 The Shortcut Worth Teaching Well queries
What is the integral of $$e^{2x}$$?
The integral of $$e^{2x}$$ is $$\tfrac{1}{2}e^{2x}+C$$.
Why is there a one-half?
The one-half appears because the derivative of $$2x$$ is 2, and integration reverses that multiplication.
Can I use substitution?
Yes. Letting $$u=2x$$ is the standard method, and it leads directly to the same result.
