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Do questions 31,33, and 35 show all the works please

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.

Explore detailed solutions to find the area between the graph of functions and the x-axis for improper integrals in questions 31, 33, and 35. Topics include convergence and divergence of integrals, symmetry of functions, and odd function properties. This guide is suitable for college-level calculus.

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