Respuesta :
Answer:
Converges at -1
Step-by-step explanation:
The integral converges if the limit exists, if the limit does not exist or if the limit is infinity it diverges.
We will make use of integral by parts to determine:
let:
[tex]u=x[/tex] [tex]dv=e^(2x)\cdot{dx}[/tex]
[tex]du=dx[/tex] [tex]v=2\cdot{e^(2x)}[/tex]
[tex]\int\limits^a_b {u} \, dv = uv -\int\limits^a_b {v} \, du[/tex]
[tex]\int\limits^a_b {x\cdot{e^2^x} \, dx =2xe^2^x- \int\limits^a_b {2e^2^x} \, dx[/tex]
[tex]\int\limits^a_b {xe^2^x} \, dx = 2xe^2^x-2e^2^x-C[/tex]
We can therefore determine that if x tends to 0 the limit is -1
[tex]\lim_{x \to \0} 2xe^2^x-2e^2^x=0-1=-1[/tex]
Answer:
The improper integral converges.
[tex]\displaystyle \int\limits^0_{- \infty} {xe^{2x}} \, dx = \frac{-1}{4}[/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]
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 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:
*Note:
When using integration by parts, ignore the integration constant C when finding v.
Step 1: Define
Identify.
[tex]\displaystyle \int\limits^0_{- \infty} {xe^{2x}} \, dx[/tex]
Step 2: Integrate Pt. 1
- [Integral] Rewrite [Improper Integral]: [tex]\displaystyle \int\limits^0_{- \infty} {xe^{2x}} \, dx = \lim_{a \to - \infty} \int\limits^0_{a} {xe^{2x}} \, 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^{2x} \ dx[/tex]
Find v using u-substitution.
- [dv] Set z: [tex]\displaystyle z = 2x[/tex]
- [z] Differentiate [Derivative Properties and Rules]: [tex]\displaystyle dz = 2 \ dx[/tex]
- [dv] Rewrite: [tex]\displaystyle \int {} \, dv = \int {e^{2x}} \, dx[/tex]
- [dv] Apply Integration Rule [Reverse Power Rule]: [tex]\displaystyle v = \int {e^{2x}} \, dx[/tex]
- Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle v = \frac{1}{2} \int {2e^{2x}} \, dx[/tex]
- [Integral] Apply Integration Method [U-Substitution]: [tex]\displaystyle v = \frac{1}{2} \int {e^{z}} \, dz[/tex]
- [v] Apply Exponential Integration: [tex]\displaystyle v = \frac{e^z}{2}[/tex]
- [z] Back-substitute: [tex]\displaystyle v = \frac{e^{2x}}{2}[/tex]
Step 3: Integrate Pt. 3
- [Integral] Apply Integration by Parts: [tex]\displaystyle \int\limits^0_{- \infty} {xe^{2x}} \, dx = \lim_{a \to - \infty} \Bigg[ \frac{xe^{2x}}{2} \bigg| \limits^0_a - \int\limits^0_a {\frac{e^{2x}}{2}} \, dx \Bigg][/tex]
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int\limits^0_{- \infty} {xe^{2x}} \, dx = \lim_{a \to - \infty} \Bigg[ \frac{xe^{2x}}{2} \bigg| \limits^0_a - \frac{1}{4} \int\limits^0_a {2e^{2x}} \, dx \Bigg][/tex]
- [Integral] Apply U-Substitution: [tex]\displaystyle \int\limits^0_{- \infty} {xe^{2x}} \, dx = \lim_{a \to - \infty} \Bigg[ \frac{xe^{2x}}{2} \bigg| \limits^0_a - \frac{1}{4} \int\limits^{x = 0}_{x = a} {e^{z}} \, dz \Bigg][/tex]
- [Integral] Apply Exponential Integration: [tex]\displaystyle \int\limits^0_{- \infty} {xe^{2x}} \, dx = \lim_{a \to - \infty} \Bigg[ \frac{xe^{2x}}{2} \bigg| \limits^0_a - \frac{1}{4} \bigg( e^z \bigg) \bigg| \limits^{x = 0}_{x = a} \Bigg][/tex]
- [z] Back-substitute: [tex]\displaystyle \int\limits^0_{- \infty} {xe^{2x}} \, dx = \lim_{a \to - \infty} \Bigg[ \frac{xe^{2x}}{2} \bigg| \limits^0_a - \frac{1}{4} \bigg( e^{2x} \bigg) \bigg| \limits^{0}_{a} \Bigg][/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^0_{- \infty} {xe^{2x}} \, dx = \lim_{a \to - \infty} \Bigg( \frac{-ae^{2a}}{2} - \frac{1 - e^{2a}}{4} \Bigg)[/tex]
- Simplify: [tex]\displaystyle \int\limits^0_{- \infty} {xe^{2x}} \, dx = \lim_{a \to - \infty} e^{2a} \bigg( \frac{1}{4} - \frac{a}{2} \bigg) - \frac{1}{4}[/tex]
- [Limit] Evaluate [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \int\limits^0_{- \infty} {xe^{2x}} \, dx = e^{2(- \infty)} \bigg( \frac{1}{4} - \frac{- \infty}{2} \bigg) - \frac{1}{4}[/tex]
- Simplify: [tex]\displaystyle \int\limits^0_{- \infty} {xe^{2x}} \, dx = - \frac{1}{4}[/tex]
∴ the improper integral equals to -0.25 and is convergent.
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Learn more about improper integrals: https://brainly.com/question/14413973
Learn more about calculus: https://brainly.com/question/23558817
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Topic: AP Calculus BC (Calculus I + II)
Unit: Integration
