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.

Learn how to evaluate the integral of the exponential function 5e^(-2x) with respect to x. This solution uses key calculus concepts and the integral formula for exponential functions, suitable for advanced high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation