Math Problem Statement

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

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

Solution

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. ∫3∞1x2 dx\int_{3}^{\infty} \frac{1}{x^2} \, dx

  1. Set up the integral as a limit: ∫3∞1x2 dx=lim⁡b→∞∫3b1x2 dx\int_{3}^{\infty} \frac{1}{x^2} \, dx = \lim_{b \to \infty} \int_{3}^{b} \frac{1}{x^2} \, dx

  2. Evaluate the integral: ∫1x2 dx=−1x\int \frac{1}{x^2} \, dx = -\frac{1}{x} So, ∫3b1x2 dx=[−1x]3b=−1b+13\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→∞b \to \infty: lim⁡b→∞(−1b+13)=0+13=13\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 13\frac{1}{3}.

3. ∫4∞2x dx\int_{4}^{\infty} \frac{2}{\sqrt{x}} \, dx

  1. Set up the integral as a limit: ∫4∞2x dx=lim⁡b→∞∫4b2x dx\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: ∫2x dx=∫2x−12 dx=4x\int \frac{2}{\sqrt{x}} \, dx = \int 2x^{-\frac{1}{2}} \, dx = 4 \sqrt{x} So, ∫4b2x dx=[4x]4b=4b−44\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→∞b \to \infty: lim⁡b→∞(4b−4⋅2)=lim⁡b→∞(4b−8)\lim_{b \to \infty} \left( 4 \sqrt{b} - 4 \cdot 2 \right) = \lim_{b \to \infty} \left( 4 \sqrt{b} - 8 \right) Since 4b4 \sqrt{b} diverges as b→∞b \to \infty, the integral diverges.

Conclusion: The integral diverges.

7. ∫1∞1x1.0001 dx\int_{1}^{\infty} \frac{1}{x^{1.0001}} \, dx

  1. Set up the integral as a limit: ∫1∞1x1.0001 dx=lim⁡b→∞∫1bx−1.0001 dx\int_{1}^{\infty} \frac{1}{x^{1.0001}} \, dx = \lim_{b \to \infty} \int_{1}^{b} x^{-1.0001} \, dx

  2. Evaluate the integral: ∫x−1.0001 dx=x−0.0001−0.0001=−x−0.00010.0001\int x^{-1.0001} \, dx = \frac{x^{-0.0001}}{-0.0001} = -\frac{x^{-0.0001}}{0.0001} So, ∫1bx−1.0001 dx=[−x−0.00010.0001]1b=−b−0.00010.0001+10.0001\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→∞b \to \infty: lim⁡b→∞(−b−0.00010.0001+10.0001)=0+10.0001=10000\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 1000010000.

9. ∫−10−2x−2 dx\int_{-10}^{-2} x^{-2} \, dx

  1. Set up the integral as a limit (since x=−2x = -2 is a discontinuity): ∫−10−2x−2 dx=lim⁡a→−2−∫−10ax−2 dx\int_{-10}^{-2} x^{-2} \, dx = \lim_{a \to -2^{-}} \int_{-10}^{a} x^{-2} \, dx

  2. Evaluate the integral: ∫x−2 dx=−1x\int x^{-2} \, dx = -\frac{1}{x} So, ∫−10ax−2 dx=[−1x]−10a=−1a+1−10\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→−2−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

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Math Problem Analysis

Mathematical Concepts

Calculus
Improper Integrals
Limits

Formulas

Integral of x^n: ∫ x^n dx = (x^(n+1)) / (n+1) + C, for n ≠ -1
Improper integral convergence: ∫ f(x) dx from a to ∞ converges if lim(b→∞) ∫ from a to b f(x) dx is finite

Theorems

Convergence of Improper Integrals
Limit Properties

Suitable Grade Level

College Level

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