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.