Respuesta :
Answer:
The improper integral converges.
[tex]\displaystyle \int\limits^{-27}_{- \infty} {x^\Big{\frac{-5}{3}}} \, dx = \frac{-1}{6}[/tex]
General Formulas and Concepts:
Calculus
Limit
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
Integration
- Integrals
Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Rule [Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
Improper Integral: [tex]\displaystyle \int\limits^{\infty}_a {f(x)} \, dx = \lim_{b \to \infty} \int\limits^b_a {f(x)} \, dx[/tex]
Step-by-step explanation:
Step 1: Define
Identify.
[tex]\displaystyle \int\limits^{-27}_{- \infty} {x^\Big{\frac{-5}{3}}} \, dx[/tex]
Step 2: Integrate
- [Integral] Rewrite [Improper Integral]: [tex]\displaystyle \int\limits^{-27}_{- \infty} {x^\Big{\frac{-5}{3}}} \, dx = \lim_{a \to - \infty} \int\limits^{-27}_{a} {x^\Big{\frac{-5}{3}}} \, dx[/tex]
- [Integral] Apply Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int\limits^{-27}_{- \infty} {x^\Big{\frac{-5}{3}}} \, dx = \lim_{a \to - \infty} \frac{-3}{2x^\big{\frac{2}{3}}} \bigg| \limits^{-27}_{a}[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^{-27}_{- \infty} {x^\Big{\frac{-5}{3}}} \, dx = \lim_{a \to - \infty} \bigg( \frac{3}{2a^\big{\frac{2}{3}}} - \frac{1}{6} \bigg)[/tex]
- [Limit] Evaluate [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \int\limits^{-27}_{- \infty} {x^\Big{\frac{-5}{3}}} \, dx = \frac{3}{2(- \infty)^\big{\frac{2}{3}}} - \frac{1}{6}[/tex]
- Simplify: [tex]\displaystyle \int\limits^{-27}_{- \infty} {x^\Big{\frac{-5}{3}}} \, dx = - \frac{1}{6}[/tex]
∴ the improper integral equals [tex]\displaystyle \bold{\frac{-1}{6}}[/tex] and is convergent.
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Learn more about improper integrals: https://brainly.com/question/14411716
Learn more about calculus: https://brainly.com/question/23558817
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Topic: AP Calculus BC (Calculus I + II)
Unit: Integration
