The image contains two integration problems from an in-class activity labeled "In-Class Act 15" for a course called "Math 2045-07."

The two integrals are as follows:

(a)
$\int_{2}^{\infty} x^{-3} \, dx$

(b)
$\int_{-\infty}^{\infty} x^{-3} \, dx$

Let's solve these integrals.

Problem (a):

$\int_{2}^{\infty} x^{-3} \, dx$

To solve this, we integrate $x^{-3}$:

1. $x^{-3}$ can be rewritten as $x^{-n}$ where $n = 3$. The integral of $x^{-n}$ is: $\int x^{-n} \, dx = \frac{x^{-n+1}}{-n+1} = \frac{x^{-2}}{-2}$

2. Apply the limits $2$ to $\infty$: $\left[ \frac{-1}{2x^2} \right]_{2}^{\infty}$

At $x = \infty$, $\frac{1}{x^2}$ tends to 0. So: $\frac{-1}{2(\infty)^2} = 0$

At $x = 2$: $\frac{-1}{2(2^2)} = \frac{-1}{8}$

Thus, the result is: $0 - \left( \frac{-1}{8} \right) = \frac{1}{8}$

So, the answer to part (a) is $\frac{1}{8}$.

Problem (b):

$\int_{-\infty}^{\infty} x^{-3} \, dx$

For this integral, notice that the function $x^{-3}$ is not defined at $x = 0$, and the integral from $-\infty$ to $\infty$ is improper. When attempting to split this integral, the portion near 0 diverges. Specifically:

1. For $x > 0$, the integral converges, similar to part (a).
2. For $x < 0$, the behavior is symmetric but opposite, and $x^{-3}$ behaves similarly. However, when the integrand is extended to 0, the integral does not converge because of the vertical asymptote at $x = 0$.

Thus, the integral diverges due to the singularity at $x = 0$.

Would you like further details on any of these steps?

Here are five related questions to expand the topic:

1. How do you determine whether an improper integral converges or diverges?
2. What methods can be used to evaluate integrals over infinite limits?
3. How do singularities affect the evaluation of integrals?
4. What is the difference between divergent and convergent improper integrals?
5. How does the comparison test work for improper integrals?

Tip: Always check the behavior of an integrand at points of discontinuity or infinity when dealing with improper integrals!