Respuesta :
Answer:
[tex] \int_{-\infty}^0 5 e^{60x} dx = \frac{1}{12}[e^0 -0]= \frac{1}{12}[/tex]
Step-by-step explanation:
Assuming this integral:
[tex] \int_{-\infty}^0 5 e^{60x} dx[/tex]
We can do this as the first step:
[tex] 5 \int_{-\infty}^0 e^{60x} dx[/tex]
Now we can solve the integral and we got:
[tex] 5 \frac{e^{60x}}{60} \Big|_{-\infty}^0 [/tex]
[tex] \int_{-\infty}^0 5 e^{60x} dx = \frac{e^{60x}}{12}\Big|_{-\infty}^0 = \frac{1}{12} [e^{60*0} -e^{-\infty}][/tex]
[tex] \int_{-\infty}^0 5 e^{60x} dx = \frac{1}{12}[e^0 -0]= \frac{1}{12}[/tex]
So then we see that the integral on this case converges amd the values is 1/12 on this case.
Answer:
The improper integral converges.
[tex]\displaystyle \int\limits^0_{- \infty} {5e^{60x}} \, dx = \frac{1}{12}[/tex]
General Formulas and Concepts:
Calculus
Limit
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 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^0_{- \infty} {5e^{60x}} \, dx[/tex]
Step 2: Integrate Pt. 1
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int\limits^0_{- \infty} {5e^{60x}} \, dx = 5 \int\limits^0_{- \infty} {e^{60x}} \, dx[/tex]
- [Integral] Rewrite [Improper Integral]: [tex]\displaystyle \int\limits^0_{- \infty} {5e^{60x}} \, dx = \lim_{a \to - \infty} 5 \int\limits^0_{a} {e^{60x}} \, dx[/tex]
Step 3: Integrate Pt. 2
Identify variables for u-substitution.
- Set u: [tex]\displaystyle u = 60x[/tex]
- [u] Differentiate [Derivative Properties and Rules]: [tex]\displaystyle du = 60 \ dx[/tex]
- [Bounds] Swap: [tex]\displaystyle \left \{ {{x = 0 \rightarrow u = 0} \atop {x = a \rightarrow u = 60a}} \right.[/tex]
Step 4: Integrate Pt. 3
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int\limits^0_{- \infty} {5e^{60x}} \, dx = \lim_{a \to - \infty} \frac{1}{12} \int\limits^0_{a} {60e^{60x}} \, dx[/tex]
- [Integral] Apply Integration Method [U-Substitution]: [tex]\displaystyle \int\limits^0_{- \infty} {5e^{60x}} \, dx = \lim_{a \to - \infty} \frac{1}{12} \int\limits^0_{60a} {e^{u}} \, du[/tex]
- [Integral] Apply Exponential Integration: [tex]\displaystyle \int\limits^0_{- \infty} {5e^{60x}} \, dx = \lim_{a \to - \infty} \frac{1}{12} e^u \bigg| \limits^0_{60a}[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^0_{- \infty} {5e^{60x}} \, dx = \lim_{a \to - \infty} \frac{1 - e^{60a}}{12}[/tex]
- [Limit] Evaluate [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle \int\limits^0_{- \infty} {5e^{60x}} \, dx = \frac{1 - e^{60(-\infty)}}{12}[/tex]
- Rewrite: [tex]\displaystyle \int\limits^0_{- \infty} {5e^{60x}} \, dx = \frac{1}{12} - \frac{1}{12e^{60(\infty)}}[/tex]
- Simplify: [tex]\displaystyle \int\limits^0_{- \infty} {5e^{60x}} \, dx = \frac{1}{12}[/tex]
∴ the improper integral equals [tex]\displaystyle \bold{\frac{1}{12}}[/tex] and is convergent.
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Learn more about improper integrals: https://brainly.com/question/14413972
Learn more about calculus: https://brainly.com/question/23558817
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Topic: AP Calculus BC (Calculus I + II)
Unit: Integration
