The problem requires finding the power series representation of the function $f(x)$ using known series expansions. The functions are given as:

$v) \quad f(x) = \int_0^1 e^{-x^2} - 1 \, dx$ $w) \quad f(x) = \int_0^x e^{-t^2} \, dt$

Let's begin with $v)$ and $w)$.

For $v)$:

We need to express $f(x) = \int_0^1 e^{-x^2} - 1 \, dx$ using a series representation for $e^{-x^2}$. The Taylor series for $e^{-x^2}$ around $x = 0$ is:

$e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$

Thus, we can rewrite the integrand $e^{-x^2} - 1$ as:

$e^{-x^2} - 1 = \sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$

Now, we integrate term by term from 0 to 1:

$f(x) = \int_0^1 \sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!} \, dx$

This can be integrated for each term in the series.

For $w)$:

The function $f(x) = \int_0^x e^{-t^2} \, dt$ involves the Gaussian integral, which does not have a simple closed-form solution. However, we can still expand $e^{-t^2}$ in a power series as before:

$e^{-t^2} = \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{n!}$

Substituting this into the integral:

$f(x) = \int_0^x \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{n!} \, dt$

Now integrate term by term:

$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_0^x t^{2n} \, dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \cdot \frac{x^{2n+1}}{2n+1}$

This is the power series representation for $f(x)$.

Would you like a more detailed step-by-step solution for the integrals or clarifications on specific parts?

Here are some follow-up questions to consider:

1. Can you express the final power series for $f(x)$ in a compact form?
2. How can you verify the convergence of the series representations?
3. What is the behavior of $f(x)$ as $x \to 0$ and $x \to \infty$?
4. How does the series for $e^{-x^2}$ compare to other known expansions, such as sine or cosine?
5. Can you use the power series to approximate $f(x)$ for small $x$?

Tip: Power series can be a very effective way to approximate functions, especially when the exact integral is difficult to compute.