Respuesta :
Answer:
INtegral converges to 1
Step-by-step explanation:
Given is an improper integral and limits are x=0 to infinity
f(x) = [tex]\frac{1}{(x+1)^2}[/tex]
is to be integrated
f(x) is defined for the values from 0 to infinity since only at -1 f becomes undefined.
We have
[tex]\int \frac{1}{(x+1)^2}dx=-\frac{1}{(x+1)[tex]
Substitute limits
When we substitute infinity this becomes 0
Hence integral value is = 0-(-1/1)=1
INtegral converges to 1
Answer:
The improper integral converges.
[tex]\displaystyle \int\limits^{\infty}_0 {\frac{dx}{(x + 1)^2}} = 1[/tex]
General Formulas and Concepts:
Calculus
Limits
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Derivative Rule [Basic Power Rule]:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
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]
{f(x)} \, dx
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Integration Property [Addition/Subtraction]: [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
Integration Method: U-Substitution
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}_0 {\frac{dx}{(x + 1)^2}}[/tex]
Step 2: Integrate Pt. 1
- [Integral] Rewrite [Improper Integral]: [tex]\displaystyle \int\limits^{\infty}_0 {\frac{dx}{(x + 1)^2}} = \lim_{b \to \infty} \int\limits^{b}_0 {\frac{dx}{(x + 1)^2}}[/tex]
Step 3: Integrate Pt. 2
Identify variables for u-substitution.
- Set u: [tex]\displaystyle u = x + 1[/tex]
- [u] Differentiate [Derivative Properties and Rules]: [tex]\displaystyle du = dx[/tex]
- [Bounds] Swap: [tex]\displaystyle \left \{ {{x = b \rightarrow u = b + 1} \atop {x = 0 \rightarrow u = 1}} \right.[/tex]
Step 4: Integrate Pt. 3
- [Integral] Apply Integration Method [U-Substitution]: [tex]\displaystyle \int\limits^{\infty}_0 {\frac{dx}{(x + 1)^2}} = \lim_{b \to \infty} \int\limits^{b + 1}_1 {\frac{du}{u^2}}[/tex]
- [Integral] Apply Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int\limits^{\infty}_0 {\frac{dx}{(x + 1)^2}} = \lim_{b \to \infty} \frac{-1}{u} \bigg| \limits^{b + 1}_1[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^{\infty}_0 {\frac{dx}{(x + 1)^2}} = \lim_{b \to \infty} \bigg( 1 - \frac{1}{b + 1} \bigg)[/tex]
- [Limit] Evaluate [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \int\limits^{\infty}_0 {\frac{dx}{(x + 1)^2}} = 1 - \frac{1}{\infty + 1}[/tex]
- Simplify: [tex]\displaystyle \int\limits^{\infty}_0 {\frac{dx}{(x + 1)^2}} = 1[/tex]
∴ the improper integral is equal to 1 and is convergent.
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Learn more about improper integrals: https://brainly.com/question/14412496
Learn more about calculus: https://brainly.com/question/20197752
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Topic: AP Calculus BC (Calculus I + II)
Unit: Integration
