Respuesta :
Answer:converge at [tex]I=\frac{1}{3}[/tex]
Step-by-step explanation:
Given
Improper Integral I is given as
[tex]I=\int^{\infty}_{3}\frac{1}{x^2}dx[/tex]
integration of [tex]\frac{1}{x^2}[/tex] is -[tex]\frac{1}{x}[/tex]
[tex]I=\left [ -\frac{1}{x}\right ]^{\infty}_3[/tex]
substituting value
[tex]I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ][/tex]
[tex]I=-\left [ 0-\frac{1}{3}\right ][/tex]
[tex]I=\frac{1}{3}[/tex]
so the value of integral converges at [tex]\frac{1}{3}[/tex]
Answer:
The improper integral converges.
[tex]\displaystyle \int\limits^{\infty}_3 {\frac{1}{x^2}} \, dx = \frac{1}{3}[/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^{\infty}_3 {\frac{1}{x^2}} \, dx[/tex]
Step 2: Integrate
- [Integral] Rewrite [Improper Integral]: [tex]\displaystyle \int\limits^{\infty}_3 {\frac{1}{x^2}} \, dx = \lim_{b \to \infty} \int\limits^{b}_3 {\frac{1}{x^2}} \, dx[/tex]
- [Integral] Apply Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int\limits^{\infty}_3 {\frac{1}{x^2}} \, dx = \lim_{b \to \infty} \frac{-1}{x} \bigg| \limits^{b}_3[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^{\infty}_3 {\frac{1}{x^2}} \, dx = \lim_{b \to \infty} \bigg( \frac{1}{3} - \frac{1}{b} \bigg)[/tex]
- [Limit] Evaluate [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \int\limits^{\infty}_3 {\frac{1}{x^2}} \, dx = \frac{1}{3} - \frac{1}{\infty}[/tex]
- Simplify: [tex]\displaystyle \int\limits^{\infty}_3 {\frac{1}{x^2}} \, dx = \frac{1}{3}[/tex]
∴ the improper integral equals [tex]\displaystyle \bold{\frac{1}{3}}[/tex] and is convergent.
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Learn more about improper integrals: https://brainly.com/question/14412088
Learn more about calculus: https://brainly.com/question/23558817
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Topic: AP Calculus BC (Calculus I + II)
Unit: Integration
