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Do questions 1,3,7,and 9 and show the work please

Let's go through questions 1, 3, 7, and 9 step-by-step, determining if each improper integral converges or diverges, and calculating the value if it converges.

### 1. \( \int_{3}^{\infty} \frac{1}{x^2} \, dx \)

1. **Set up the integral as a limit**:
   \[
   \int_{3}^{\infty} \frac{1}{x^2} \, dx = \lim_{b \to \infty} \int_{3}^{b} \frac{1}{x^2} \, dx
   \]

2. **Evaluate the integral**:
   \[
   \int \frac{1}{x^2} \, dx = -\frac{1}{x}
   \]
   So,
   \[
   \int_{3}^{b} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_{3}^{b} = -\frac{1}{b} + \frac{1}{3}
   \]

3. **Take the limit as \( b \to \infty \)**:
   \[
   \lim_{b \to \infty} \left( -\frac{1}{b} + \frac{1}{3} \right) = 0 + \frac{1}{3} = \frac{1}{3}
   \]

**Conclusion**: The integral converges, and the value is \( \frac{1}{3} \).

---

### 3. \( \int_{4}^{\infty} \frac{2}{\sqrt{x}} \, dx \)

1. **Set up the integral as a limit**:
   \[
   \int_{4}^{\infty} \frac{2}{\sqrt{x}} \, dx = \lim_{b \to \infty} \int_{4}^{b} \frac{2}{\sqrt{x}} \, dx
   \]

2. **Rewrite and evaluate the integral**:
   \[
   \int \frac{2}{\sqrt{x}} \, dx = \int 2x^{-\frac{1}{2}} \, dx = 4 \sqrt{x}
   \]
   So,
   \[
   \int_{4}^{b} \frac{2}{\sqrt{x}} \, dx = \left[ 4 \sqrt{x} \right]_{4}^{b} = 4 \sqrt{b} - 4 \sqrt{4}
   \]

3. **Take the limit as \( b \to \infty \)**:
   \[
   \lim_{b \to \infty} \left( 4 \sqrt{b} - 4 \cdot 2 \right) = \lim_{b \to \infty} \left( 4 \sqrt{b} - 8 \right)
   \]
   Since \( 4 \sqrt{b} \) diverges as \( b \to \infty \), the integral diverges.

**Conclusion**: The integral diverges.

---

### 7. \( \int_{1}^{\infty} \frac{1}{x^{1.0001}} \, dx \)

1. **Set up the integral as a limit**:
   \[
   \int_{1}^{\infty} \frac{1}{x^{1.0001}} \, dx = \lim_{b \to \infty} \int_{1}^{b} x^{-1.0001} \, dx
   \]

2. **Evaluate the integral**:
   \[
   \int x^{-1.0001} \, dx = \frac{x^{-0.0001}}{-0.0001} = -\frac{x^{-0.0001}}{0.0001}
   \]
   So,
   \[
   \int_{1}^{b} x^{-1.0001} \, dx = \left[ -\frac{x^{-0.0001}}{0.0001} \right]_{1}^{b} = -\frac{b^{-0.0001}}{0.0001} + \frac{1}{0.0001}
   \]

3. **Take the limit as \( b \to \infty \)**:
   \[
   \lim_{b \to \infty} \left( -\frac{b^{-0.0001}}{0.0001} + \frac{1}{0.0001} \right) = 0 + \frac{1}{0.0001} = 10000
   \]

**Conclusion**: The integral converges, and the value is \( 10000 \).

---

### 9. \( \int_{-10}^{-2} x^{-2} \, dx \)

1. **Set up the integral as a limit** (since \( x = -2 \) is a discontinuity):
   \[
   \int_{-10}^{-2} x^{-2} \, dx = \lim_{a \to -2^{-}} \int_{-10}^{a} x^{-2} \, dx
   \]

2. **Evaluate the integral**:
   \[
   \int x^{-2} \, dx = -\frac{1}{x}
   \]
   So,
   \[
   \int_{-10}^{a} x^{-2} \, dx = \left[ -\frac{1}{x} \right]_{-10}^{a} = -\frac{1}{a} + \frac{1}{-10}
   \]

3. **Take the limit as \( a \to -2^{-} \)**:
   \[
   \lim_{a \to -2^{-}} \left( -\frac{1}{a} + \frac{1}{-10} \right) = \lim_{a \to -2^{-}} \left( -\frac{1}{a} - \frac{1}{10} \right) = \frac{1}{2} - \frac{1}{10} = \frac{5}{10} - \frac{1

Do questions 1, 3, 7, and 9 and show the work please

Detailed solution on evaluating the convergence or divergence of improper integrals for questions 1, 3, 7, and 9. Learn how to set up limits for integrals extending to infinity, identify divergences, and calculate values for convergent integrals. Suitable for college-level calculus students.

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