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To find the Maclaurin series for \( f(x) = \int_0^x \left( e^{-8t^5} - 1 \right) \, dt \), we can follow these steps:

1. **Expand \( e^{-8t^5} \) as a Taylor series** around \( t = 0 \) (i.e., a Maclaurin series) for easier integration. The Maclaurin series for \( e^u \) is:
   \[
   e^u = \sum_{k=0}^{\infty} \frac{u^k}{k!}
   \]
   In this case, \( u = -8t^5 \), so
   \[
   e^{-8t^5} = \sum_{k=0}^{\infty} \frac{(-8t^5)^k}{k!} = \sum_{k=0}^{\infty} \frac{(-8)^k t^{5k}}{k!}.
   \]

2. **Subtract 1** from this series, as specified by the integrand \( e^{-8t^5} - 1 \):
   \[
   e^{-8t^5} - 1 = \sum_{k=0}^{\infty} \frac{(-8)^k t^{5k}}{k!} - 1.
   \]
   This expression simplifies to:
   \[
   e^{-8t^5} - 1 = \sum_{k=1}^{\infty} \frac{(-8)^k t^{5k}}{k!}.
   \]

3. **Integrate term by term** with respect to \( t \) from \( 0 \) to \( x \):
   \[
   f(x) = \int_0^x \left( \sum_{k=1}^{\infty} \frac{(-8)^k t^{5k}}{k!} \right) \, dt = \sum_{k=1}^{\infty} \frac{(-8)^k}{k!} \int_0^x t^{5k} \, dt.
   \]
   Now, integrate \( t^{5k} \) with respect to \( t \):
   \[
   \int_0^x t^{5k} \, dt = \frac{x^{5k+1}}{5k+1}.
   \]
   Therefore,
   \[
   f(x) = \sum_{k=1}^{\infty} \frac{(-8)^k x^{5k+1}}{k!(5k+1)}.
   \]

Thus, the Maclaurin series for \( f(x) \) is:
\[
f(x) = \sum_{k=1}^{\infty} \frac{(-8)^k x^{5k+1}}{k!(5k+1)}.
\]

Would you like more details or have any questions about this solution?

### Related Questions
1. How do we determine the radius of convergence for this series?
2. How would the series change if the exponent of \( t \) were different (e.g., \( t^4 \) instead of \( t^5 \))?
3. What is the significance of using the Maclaurin series for analyzing functions near \( x = 0 \)?
4. Can we approximate \( f(x) \) for small values of \( x \) by using just a few terms of the series?
5. How does the convergence of this series compare with that of other Taylor series?

### Tip
When dealing with integrals of exponential functions, expanding the exponential term into a series often simplifies the integral, making it easier to evaluate term by term.

Find a Maclaurin series for \( f(x) = \int_0^x (e^{-8t^5} - 1) \, dt \).

Explore the step-by-step solution to finding the Maclaurin series for \( f(x) = \int_0^x (e^{-8t^5} - 1) \, dt \). This includes expanding \( e^{-8t^5} \) using the Taylor series, subtracting 1, and integrating term by term. Ideal for college-level calculus students.

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