Respuesta :
Answer:
The improper integral diverges.
[tex]\displaystyle \int\limits^{\infty}_0 {xe^{4x}} \, 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 Method: U-Substitution
Integration by Parts: [tex]\displaystyle \int {u} \, dv = uv - \int {v} \, du[/tex]
- [IBP] LIPET: Logs, Inverses, Polynomials, Exponentials, Trig
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:
When doing integration by parts and using LIPET, ignore the integration constant C when finding v.
Step 1: Define
Identify.
[tex]\displaystyle \int\limits^{\infty}_0 {xe^{4x}} \, dx[/tex]
Step 2: Integrate Pt. 1
- [Integral] Rewrite [Improper Integral]: [tex]\displaystyle \int\limits^{\infty}_0 {xe^{4x}} \, dx = \lim_{b \to \infty} \int\limits^b_0 {xe^{4x}} \, dx[/tex]
Step 3: Integrate Pt. 2
Identify variables for integration by parts.
- Set u [LIPET]: [tex]\displaystyle u = x[/tex]
- [u] Differentiate [Derivative Rule - Basic Power Rule]: [tex]\displaystyle du = dx[/tex]
- Set dv [LIPET]: [tex]\displaystyle dv = e^{4x} \ dx[/tex]
Find v using u-substitution.
- [dv] Set z: [tex]\displaystyle z = 4x[/tex]
- [z] Differentiate [Derivative Properties and Rules]: [tex]\displaystyle dz = 4 \ dx[/tex]
- [dv] Rewrite: [tex]\displaystyle \int {} \, dv = \int {e^{4x}} \, dx[/tex]
- [dv] Apply Integration Rule [Reverse Power Rule]: [tex]\displaystyle v = \int {e^{4x}} \, dx[/tex]
- Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle v = \frac{1}{4} \int {4e^{4x}} \, dx[/tex]
- [Integral] U-Substitution: [tex]\displaystyle v = \frac{1}{4} \int {e^u} \, du[/tex]
- [v] Apply Exponential Integration: [tex]\displaystyle v = \frac{e^u}{4}[/tex]
- [u] Back-substitute: [tex]\displaystyle v = \frac{e^{4x}}{4}[/tex]
Step 4: Integrate Pt. 3
- [Integral] Apply Integration by Parts: [tex]\displaystyle \int\limits^{\infty}_0 {xe^{4x}} \, dx = \lim_{b \to \infty} \Bigg[ \frac{xe^{4x}}{4} \bigg| \limits^b_0 - \int\limits^b_0 {\frac{e^{4x}}{4}} \, dx \Bigg][/tex]
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int\limits^{\infty}_0 {xe^{4x}} \, dx = \lim_{b \to \infty} \Bigg[ \frac{xe^{4x}}{4} \bigg| \limits^b_0 - \frac{1}{16} \int\limits^b_0 {4e^{4x} \, dx \Bigg][/tex]
- [Integral] Apply U-Substitution: [tex]\displaystyle \int\limits^{\infty}_0 {xe^{4x}} \, dx = \lim_{b \to \infty} \Bigg[ \frac{xe^{4x}}{4} \bigg| \limits^b_0 - \frac{1}{16} \int\limits^b_0 {e^{u} \, du \Bigg][/tex]
- [Integral] Apply Exponential Integration: [tex]\displaystyle \int\limits^{\infty}_0 {xe^{4x}} \, dx = \lim_{b \to \infty} \Bigg[ \frac{xe^{4x}}{4} \bigg| \limits^b_0 - \frac{1}{16} \bigg( e^u \bigg) \bigg| \limits^{x = b}_{x = 0} \Bigg][/tex]
- [u] Back-substitute: [tex]\displaystyle \int\limits^{\infty}_0 {xe^{4x}} \, dx = \lim_{b \to \infty} \Bigg[ \frac{xe^{4x}}{4} \bigg| \limits^b_0 - \frac{1}{16} \bigg( e^{4x} \bigg) \bigg| \limits^b_0 \Bigg][/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^{\infty}_0 {xe^{4x}} \, dx = \lim_{b \to \infty} \Bigg[ \frac{be^{4b}}{4} - \frac{1}{16} \bigg( e^{4b} - 1 \bigg) \Bigg][/tex]
- Simplify: [tex]\displaystyle \int\limits^{\infty}_0 {xe^{4x}} \, dx = \lim_{b \to \infty} e^{4b} \bigg( \frac{b}{4} - \frac{1}{16} \bigg) + \frac{1}{16}[/tex]
- [Limit] Evaluate [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \int\limits^{\infty}_0 {xe^{4x}} \, dx = e^{4(\infty)} \bigg( \frac{\infty}{4} - \frac{1}{16} \bigg) + \frac{1}{16}[/tex]
- Simplify: [tex]\displaystyle \int\limits^{\infty}_0 {xe^{4x}} \, dx = \infty[/tex]
∴ the improper integral tends to ∞ and is divergent.
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Learn more about improper integrals: https://brainly.com/question/14412891
Learn more about calculus: brainly.com/question/23558817
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Topic: AP Calculus BC (Calculus I + II)
Unit: Integration
