To evaluate the integral

$\int 5e^{-2x} \, dx,$

we can proceed by recognizing that this is an exponential function, which integrates by a basic rule.

1. Factor out the constant 5: $\int 5e^{-2x} \, dx = 5 \int e^{-2x} \, dx.$

2. Integrate $e^{-2x}$: We use the rule for integrating $e^{ax}$: $\int e^{ax} \, dx = \frac{e^{ax}}{a} + C.$ Here, $a = -2$, so $\int e^{-2x} \, dx = \frac{e^{-2x}}{-2} = -\frac{1}{2}e^{-2x}.$

3. Multiply by the constant 5: $5 \int e^{-2x} \, dx = 5 \left(-\frac{1}{2} e^{-2x}\right) = -\frac{5}{2} e^{-2x}.$

4. Add the constant of integration $C$: $\int 5e^{-2x} \, dx = -\frac{5}{2} e^{-2x} + C.$

Final Answer:

$\int 5e^{-2x} \, dx = -\frac{5}{2} e^{-2x} + C.$

Would you like further details on this?

Here are some related questions:

1. How do you integrate functions of the form $e^{ax}$?
2. What is the integral of $e^{kx} \sin x$ or $e^{kx} \cos x$?
3. Can you explain the rule for integrating $e^{ax + b}$?
4. What is the method to solve definite integrals of $e^{ax}$ over an interval?
5. How does the derivative of $e^{ax}$ relate to its integral?

Tip: When integrating exponential functions, remember that the coefficient of $x$ in the exponent affects the result by dividing it by that coefficient.