Evaluate the integral: I = integral e^(- 5/2 x) dx
Evaluate the integral: $$ I = \int e^{-\frac{5}{2}x} \, dx. $$
Explanation of Method
The integral of $e^{ax}$ for any nonzero constant $a$ (integer, fractional, or negative) follows the same simple rule: $\int e^{ax}\,dx = \frac1a e^{ax}+C$. Here $a=-\tfrac52$, so the antiderivative is scaled by $\tfrac{1}{-5/2}=-\tfrac25$.
- Recall the general antiderivative formula for $e^{ax}$.
- Substitute the given exponent's coefficient for $a$.
- Simplify the resulting constant and add the constant of integration.
Step 1 · Recall the formula for the integral of an exponential function
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 2 · Apply the formula
In our case, the exponent $a=-\dfrac{5}{2}$. Substituting this into the formula, we have:
$$ I = \int e^{-\frac{5}{2}x}\,dx = \frac{1}{-\frac{5}{2}}e^{-\frac{5}{2}x} + C. $$
Step 3 · Simplify the result
Simplify the constant $\dfrac{1}{-\frac{5}{2}}$ as follows:
$$ \frac{1}{-\frac{5}{2}} = -\frac{2}{5}. $$
Thus, the integral becomes:
$$ I = -\frac{2}{5}e^{-\frac{5}{2}x} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $a$ | $-5/2$ |
| $1/a$ | $-2/5$ |
| Antiderivative | $-\tfrac25 e^{-5x/2}$ |
Additional Notes
- The integrand is defined for all real $x$, since $e^{ax}$ is defined everywhere for any real $a$.
- The formula $\int e^{ax}\,dx=\frac1a e^{ax}+C$ holds for any nonzero constant $a$, whether positive, negative, integer, or fractional — no special handling is needed for fractional or negative exponents beyond correctly inverting $a$.
- The result can be checked by differentiating $-\tfrac25 e^{-5x/2}$ with the chain rule and confirming it returns $e^{-5x/2}$.
Result
The solution to the integral is: