Certainly! Let's take a closer look at the integration step:

Given the Taylor series expansion: $e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$

We want to integrate: $\int_{0}^{1} e^{-x^2} \, dx = \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} \, dx$

Since the series converges uniformly on $[0, 1]$, we can interchange the sum and the integral: $\int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} \, dx = \sum_{n=0}^{\infty} \int_{0}^{1} \frac{(-1)^n x^{2n}}{n!} \, dx$

Now, let's integrate each term individually: $\int_{0}^{1} \frac{(-1)^n x^{2n}}{n!} \, dx$

This simplifies to: $\frac{(-1)^n}{n!} \int_{0}^{1} x^{2n} \, dx$

We need to evaluate the integral: $\int_{0}^{1} x^{2n} \, dx$

To integrate $x^{2n}$: $\int x^{2n} \, dx = \frac{x^{2n+1}}{2n+1} + C$

Evaluating this definite integral from 0 to 1: $\int_{0}^{1} x^{2n} \, dx = \left[ \frac{x^{2n+1}}{2n+1} \right]_{0}^{1} = \frac{1^{2n+1}}{2n+1} - \frac{0^{2n+1}}{2n+1} = \frac{1}{2n+1}$

Thus: $\int_{0}^{1} x^{2n} \, dx = \frac{1}{2n+1}$

Substituting this back into our earlier expression: $\int_{0}^{1} \frac{(-1)^n x^{2n}}{n!} \, dx = \frac{(-1)^n}{n!} \cdot \frac{1}{2n+1}$

Now, summing over all terms $n$ from 0 to $\infty$: $\int_{0}^{1} e^{-x^2} \, dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)n!}$

Therefore, the integral $\int_{0}^{1} e^{-x^2} \, dx$ written as an infinite series is: $\int_{0}^{1} e^{-x^2} \, dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)n!}$

Do you have any more questions or need further clarification?

Here are some additional questions you might consider:

1. What are some common methods for approximating definite integrals?
2. Can you provide more examples of integrating series term-by-term?
3. How do you test for uniform convergence of a series?
4. What is the significance of the factorial in the Taylor series?
5. Can you explain how the exponential function $e^{-x^2}$ is used in probability theory?

Tip: When working with series and integrals, carefully consider the conditions under which you can interchange summation and integration to ensure the steps are valid.