Evaluate the integral: I = integral (e^-x + e^-2x) dx
Evaluate the integral: $$ I = \int \left(e^{-x} + e^{-2x}\right) \, dx. $$
Explanation of Method
By linearity, the integral of a sum of exponentials with different rate constants splits into a sum of two applications of $\int e^{ax}\,dx=\frac1a e^{ax}+C$, one with $a=-1$ and one with $a=-2$.
- Split the integral into two terms using linearity.
- Apply the general exponential antiderivative formula to each term.
- Combine the results and add the constant of integration.
Step 1 · Split the integral
Using the linearity of integrals, we can split the given integral into two separate terms:
$$ I = \int e^{-x}\,dx + \int e^{-2x}\,dx. $$
Step 2 · Recall the formula for the integral of exponential functions
The integral of an exponential function of the form $\int e^{ax}\,dx$ is given by:
$$ \int e^{ax}\,dx = \frac{1}{a}e^{ax} + C, \quad \text{where } a\neq 0. $$
Step 3 · Evaluate each term
For the first term, $\displaystyle \int e^{-x}\,dx$:
$$ \int e^{-x}\,dx = \frac{1}{-1}e^{-x} = -e^{-x}. $$
For the second term, $\displaystyle \int e^{-2x}\,dx$:
$$ \int e^{-2x}\,dx = \frac{1}{-2}e^{-2x} = -\frac{1}{2}e^{-2x}. $$
Step 4 · Combine the results
Adding the two results, we get:
$$ I = -e^{-x} - \frac{1}{2}e^{-2x} + C, $$
where $C$ is the constant of integration.
Summary Table
| Term | $a$ | Antiderivative |
|---|---|---|
| $e^{-x}$ | $-1$ | $-e^{-x}$ |
| $e^{-2x}$ | $-2$ | $-\tfrac12 e^{-2x}$ |
Additional Notes
- The integrand is defined for all real $x$, since exponential functions are defined everywhere.
- Both terms decay as $x\to\infty$, but they integrate to negative antiderivatives because their rate constants are negative — the general formula $\frac1a e^{ax}$ handles the sign automatically.
- The result can be checked by differentiating $-e^{-x}-\tfrac12 e^{-2x}$ and confirming it returns $e^{-x}+e^{-2x}$.
Result
The solution to the integral is: