Integral E 2x: The Step Students Often Skip
Integral of e^(2x) Solved Clearly
The integral of e^(2x) is $$\frac{1}{2}e^{2x} + C$$. This follows from the chain rule in reverse: because the derivative of $$2x$$ is 2, the antiderivative must be divided by 2 to keep the result correct.
Why the Answer Works
For any exponential of the form $$e^{ax}$$, the general rule is $$\int e^{ax}\,dx = \frac{1}{a}e^{ax}+C$$ when $$a \neq 0$$. In this case, $$a=2$$, so the coefficient becomes $$\frac{1}{2}$$, which is why the result is not just $$e^{2x}+C$$.
| Expression | Antiderivative | Check |
|---|---|---|
| $$\int e^{2x}\,dx$$ | $$\frac{1}{2}e^{2x}+C$$ | $$\frac{d}{dx}\left(\frac{1}{2}e^{2x}\right)=e^{2x}$$ |
| $$\int e^{x}\,dx$$ | $$e^{x}+C$$ | $$\frac{d}{dx}(e^{x})=e^{x}$$ |
| $$\int e^{3x}\,dx$$ | $$\frac{1}{3}e^{3x}+C$$ | $$\frac{d}{dx}\left(\frac{1}{3}e^{3x}\right)=e^{3x}$$ |
Step-by-Step Method
- Identify the inner function $$2x$$.
- Let $$u=2x$$, so $$du=2\,dx$$.
- Rewrite $$dx=\frac{1}{2}du$$.
- Substitute into the integral: $$\int e^{2x}dx=\frac{1}{2}\int e^u\,du$$.
- Integrate to get $$\frac{1}{2}e^u+C$$.
- Replace $$u$$ with $$2x$$: $$\frac{1}{2}e^{2x}+C$$.
Common Mistakes
- Forgetting the constant $$C$$, which is required for every indefinite integral.
- Writing $$e^{2x}+C$$ without the factor $$\frac{1}{2}$$.
- Confusing the integral of $$e^{2x}$$ with the derivative of $$e^{2x}$$, which is $$2e^{2x}$$.
- Skipping the substitution check, which helps confirm the result is correct.
"The key is to match the inside derivative. If the exponent is $$2x$$, the antiderivative must compensate with $$\frac{1}{2}$$."
Teacher-Friendly Explanation
The cleanest classroom explanation is that exponential functions keep their shape under integration, but the inside coefficient must be adjusted. That makes u-substitution the standard method for explaining why $$\int e^{2x}dx=\frac{1}{2}e^{2x}+C$$.
Quick Reference
| Function | Integral |
|---|---|
| $$e^{ax}$$ | $$\frac{1}{a}e^{ax}+C$$ |
| $$e^{2x}$$ | $$\frac{1}{2}e^{2x}+C$$ |
| $$e^{-3x}$$ | $$-\frac{1}{3}e^{-3x}+C$$ |
What are the most common questions about Integral E 2x The Step Students Often Skip?
What is the integral of e^(2x)?
The integral of $$e^{2x}$$ is $$\frac{1}{2}e^{2x}+C$$.
Why is there a 1/2?
The $$\frac{1}{2}$$ corrects for the inner derivative of $$2x$$, so differentiating the result gives back the original function.
Can I use substitution?
Yes. Setting $$u=2x$$ is the standard way to solve it cleanly and verify the rule.
Is the answer exact?
Yes. $$\frac{1}{2}e^{2x}+C$$ is the exact indefinite integral, with $$C$$ representing any constant.
