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The integral given is:

\[
\int 5e^{7 - 3t} \, dt
\]

To solve this, let’s start by factoring out the constant and then addressing the exponent.

### Step-by-Step Solution
1. **Rewrite the Integral:**

   \[
   \int 5e^{7 - 3t} \, dt = 5 \int e^{7 - 3t} \, dt
   \]

2. **Substitute for Simplification:**
   Let \( u = 7 - 3t \). Then,
   \[
   \frac{du}{dt} = -3 \Rightarrow dt = -\frac{1}{3} du
   \]

3. **Substitute \( u \) and \( dt \) into the Integral:**
   \[
   5 \int e^{u} \cdot \left(-\frac{1}{3}\right) \, du = -\frac{5}{3} \int e^{u} \, du
   \]

4. **Integrate with Respect to \( u \):**
   The integral of \( e^{u} \) is \( e^{u} \), so:
   \[
   -\frac{5}{3} \int e^{u} \, du = -\frac{5}{3} e^{u}
   \]

5. **Substitute \( u = 7 - 3t \) Back:**
   \[
   -\frac{5}{3} e^{7 - 3t} + C
   \]

### Final Answer
\[
\int 5e^{7 - 3t} \, dt = -\frac{5}{3} e^{7 - 3t} + C
\]

Would you like more details on any specific part, or do you have questions?

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Here are five related questions to deepen your understanding:
1. How do you handle integrals with complex exponent functions?
2. How does the chain rule relate to integration by substitution?
3. What is the significance of the constant of integration in indefinite integrals?
4. How would you integrate \( \int e^{kt} \, dt \) for any constant \( k \)?
5. What happens if you have multiple exponent terms, such as \( e^{a + bt} \)?

**Tip:** Always look for substitutions that simplify the exponent in integrals involving exponentials; it can streamline the integration process.

Explore the step-by-step solution to the integral ∫ 5e^(7 - 3t) dt using integration by substitution. This solution covers exponential function integration and substitution techniques, suitable for undergraduate calculus students.

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