Let's solve questions 31, 33, and 35 step-by-step, calculating the area between the given functions and the x-axis over the specified intervals.

Question 31

Given: $f(x) = \frac{1}{x - 1}, \quad \text{for } (-\infty, 0]$

Goal: Find the area between $f(x)$ and the x-axis over the interval $(-\infty, 0]$.

Solution:

1. Set up the integral: The area under the curve $f(x) = \frac{1}{x - 1}$ from $-\infty$ to $0$ is represented by: $\int_{-\infty}^{0} \frac{1}{x - 1} \, dx$

2. Integrate $f(x) = \frac{1}{x - 1}$: To integrate this, let's use a substitution. Let: $u = x - 1 \Rightarrow du = dx$ When $x = -\infty$, $u \to -\infty$, and when $x = 0$, $u = -1$. The integral becomes: $\int_{-\infty}^{-1} \frac{1}{u} \, du$

3. Calculate the integral: $\int_{-\infty}^{-1} \frac{1}{u} \, du = \left[ \ln |u| \right]_{-\infty}^{-1}$ Evaluating this integral gives: $\left( \ln |-1| - \lim_{u \to -\infty} \ln |u| \right) = 0 - (-\infty) = \infty$

Conclusion: The integral diverges, so the area is infinite.

Question 33

Given: $f(x) = \frac{1}{(x - 1)^2}, \quad \text{for } (-\infty, 0]$

Goal: Find the area between $f(x)$ and the x-axis over the interval $(-\infty, 0]$.

Solution:

1. Set up the integral: The area under the curve $f(x) = \frac{1}{(x - 1)^2}$ from $-\infty$ to $0$ is represented by: $\int_{-\infty}^{0} \frac{1}{(x - 1)^2} \, dx$

2. Integrate $f(x) = \frac{1}{(x - 1)^2}$: Let's again use substitution. Let: $u = x - 1 \Rightarrow du = dx$ When $x = -\infty$, $u \to -\infty$, and when $x = 0$, $u = -1$. Thus, the integral becomes: $\int_{-\infty}^{-1} \frac{1}{u^2} \, du$

3. Calculate the integral: $\int_{-\infty}^{-1} \frac{1}{u^2} \, du = \left[ -\frac{1}{u} \right]_{-\infty}^{-1}$ Evaluating this integral: $-\frac{1}{-1} - \lim_{u \to -\infty} \left(-\frac{1}{u}\right) = 1 - 0 = 1$

Conclusion: The area is $\boxed{1}$.

Question 35

Given: $\int_{-\infty}^{\infty} x e^{-x^2} \, dx$

Goal: Evaluate the integral.

Solution:

1. Observe the symmetry of the integrand: Notice that $x e^{-x^2}$ is an odd function because: $f(-x) = -x e^{-(-x)^2} = -x e^{-x^2} = -f(x)$ Since the function is odd and we are integrating over a symmetric interval $(-\infty, \infty)$, the integral of an odd function over a symmetric interval around zero is always zero.

Conclusion: $\int_{-\infty}^{\infty} x e^{-x^2} \, dx = \boxed{0}$

Would you like any further details on these solutions, or do you have any questions?

Related Questions

1. How do you determine if an integral diverges or converges?
2. What are some common techniques for evaluating improper integrals?
3. Can all functions be integrated over infinite intervals?
4. Why do odd functions integrate to zero over symmetric intervals?
5. How do you handle integrals with asymptotes in the interval of integration?

Tip

For integrals over infinite intervals, always check if the function has symmetry properties or if it diverges at certain points to simplify your work.