Respuesta :
Answer:Divergent
Step-by-step explanation:
Given
Improper Integral I is given as
[tex]I=\int^{\infty}_{4}\frac{2}{\sqrt{x}}dx[/tex]
Integration of [tex]\frac{2}{\sqrt{x}}[/tex] is [tex]4\sqrt{2}[/tex]
[tex]I=\left [ 4\sqrt{x}\right ]^{\infty}_{4}[/tex]
[tex]I=4\left [ \sqrt{x}\right ]^{\infty}_{4}[/tex]
[tex]I=4\left [ \sqrt{\infty}-\sqrt{4}\right ][/tex]
I tends to [tex]\infty[/tex] so Integral is divergent in nature
Answer:
The improper integral diverges.
[tex]\displaystyle \int\limits^{\infty}_4 {\frac{2}{\sqrt{x}}} \, dx = \infty[/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 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]
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 Property [Splitting Integral]: [tex]\displaystyle \int\limits^c_a {f(x)} \, dx = \int\limits^b_a {f(x)} \, dx + \int\limits^c_b {f(x)} \, dx[/tex]
Integration Method: U-Substitution
Improper Integrals: [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:
*Note:
This is a doubly improper integral, as we have infinity as a bound and an infinite discontinuity at x = 4, which is another bound.
Step 1: Define
Identify.
[tex]\displaystyle \int\limits^{\infty}_4 {\frac{2}{\sqrt{x}}} \, dx[/tex]
Step 2: Integrate
- [Integral] Rewrite [Integration Property - Splitting Integral]: [tex]\displaystyle \int\limits^{\infty}_4 {\frac{2}{\sqrt{x}}} \, dx = \int\limits^{9}_4 {\frac{2}{\sqrt{x}}} \, dx + \int\limits^{\infty}_9 {\frac{2}{\sqrt{x}}} \, dx[/tex]
- [Integrals] Rewrite [Improper Integral]: [tex]\displaystyle \int\limits^{\infty}_4 {\frac{2}{\sqrt{x}}} \, dx = \lim_{a \to 4} \int\limits^{9}_a {\frac{2}{\sqrt{x}}} \, dx + \lim_{b \to \infty} \int\limits^{b}_9 {\frac{2}{\sqrt{x}}} \, dx[/tex]
- [Integrals] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int\limits^{\infty}_4 {\frac{2}{\sqrt{x}}} \, dx = \lim_{a \to 4} 2 \int\limits^{9}_a {\frac{1}{\sqrt{x}}} \, dx + \lim_{b \to \infty} 2 \int\limits^{b}_9 {\frac{1}{\sqrt{x}}} \, dx[/tex]
- [Integrals] Apply Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int\limits^{\infty}_4 {\frac{2}{\sqrt{x}}} \, dx = \lim_{a \to 4} 2 \bigg( 2\sqrt{x} \bigg) \bigg| \limits^{9}_a + \lim_{b \to \infty} 2 \bigg( 2\sqrt{x} \bigg) \bigg| \limits^{b}_9[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^{\infty}_4 {\frac{2}{\sqrt{x}}} \, dx = \lim_{a \to 4} 2 \big( 6 - 2\sqrt{a} \big) + \lim_{b \to \infty} 2 \big( 2\sqrt{b} - 6 \big)[/tex]
- Simplify: [tex]\displaystyle \int\limits^{\infty}_4 {\frac{2}{\sqrt{x}}} \, dx = \lim_{a \to 4} 4 \big( 3 - \sqrt{a} \big) + \lim_{b \to \infty} 4 \big( \sqrt{b} - 3 \big)[/tex]
- [Limits] Evaluate [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \int\limits^{\infty}_4 {\frac{2}{\sqrt{x}}} \, dx = 4 \big( 3 - \sqrt{4} \big) + 4 \big( \sqrt{\infty} - 3 \big)[/tex]
- Simplify: [tex]\displaystyle \int\limits^{\infty}_4 {\frac{2}{\sqrt{x}}} \, dx = \infty[/tex]
∴ the improper integral tends to ∞ and is divergent.
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Learn more about improper integrals: https://brainly.com/question/14414145
Learn more about calculus: brainly.com/question/23558817
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Topic: AP Calculus BC (Calculus I + II)
Unit: Integration
