For any expression $h(x)$,
let $\lim_{x \to k^+} h(x)$ denote the limit of $h(x)$ as $x$ approaches $k$, from above.
Similarly,
let $\lim_{x \to k^-} h(x)$ denote the limit of $h(x)$ as $x$ approaches $k$, from below.
b) Explain why $$\int^{-1}_{1} \frac{1}{x^2}dx$$ is undefined.
Let $f(x)$ denote $\frac{1}{x^2}$, which implies that $f(x)$ is well defined throughout $(-1,1)$, except at $x=0$.
Therefore, from the definition of improper integrals,
$$I = \int_1^{-1} f(x)dx = I_1 + I_2$$
where
$$I_1 = \lim_{b \to 0^+} \int_1^b f(x)dx$$
and
$$I_2 = \lim_{a \to 0^-} \int_a^{(-1)} f(x)dx.$$
This means that in order for $I$ to exist (i.e. be well defined) it is necessary (and sufficient) that each of $I_1$ and $I_2$ be (separately) well defined. Failure with respect to either $I_1$ or $I_2$ is sufficient to conclude that $I$ is not well defined.
Therefore, the problem reduces to showing that $I_1$ is not well defined (i.e. that the corresponding limit does not exist).
$$\int_1^b f(x)dx = F(b) - F(1)$$
where
$$F(x) = \frac{-1}{x}.$$
Therefore,
$$\int_1^b f(x)dx = \frac{-1}{b} - \frac{-1}{1} = \frac{b-1}{b}.$$
Therefore,
$$I_1 = \lim_{b \to 0^+} \frac{b-1}{b}.$$
As you can see, as $b \to 0^+,$ the above numerator goes to $-1$, and the above denominator goes to $0$, from above. Therefore, as $b \to 0^+, I_1 \to -\infty.$
Therefore, $I_1$ is not well defined.
Therefore, $I$ is not well defined.
