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Determine if the improper integral is convergent or divergent, and calculate its value if it is convergent. integral^ infinity_ 0 14 e^ -7 x dx Calculate the value of the improper integral. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. integral^ infinity_ 0 14 e^ - 7x dx The improper integral diverges.

Respuesta :

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Answer:

[tex]\displaystyle \int\limits^{\infty}_0 {14e^{-7x}} \, dx = 2[/tex]

∵ there is an integral value, the integral converges.

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]:

  1. f(x) = cxⁿ
  2. 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

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:

Step 1: Define

Identify.

[tex]\displaystyle \int\limits^{\infty}_0 {14e^{-7x}} \, dx[/tex]

Step 2: Integrate Pt. 1

  1. [Integral] Rewrite [Improper Integral]:                                                     [tex]\displaystyle \int\limits^{\infty}_0 {14e^{-7x}} \, dx = \lim_{b \to \infty} \int\limits^b_0 {14e^{-7x}} \, dx[/tex]
  2. [Integral] Rewrite [Integration Property - Multiplied Constant]:             [tex]\displaystyle \int\limits^{\infty}_0 {14e^{-7x}} \, dx = \lim_{b \to \infty} 14 \int\limits^b_0 {e^{-7x}} \, dx[/tex]

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

  1. Set u:                                                                                                         [tex]\displaystyle u = -7x[/tex]
  2. [u] Differentiate [Derivative Properties and Rules]:                                 [tex]\displaystyle du = -7 \ dx[/tex]
  3. [Bounds] Swap:                                                                                         [tex]\displaystyle \left \{ {{x = b \rightarrow u = -7b} \atop {x = 0 \rightarrow u = 0}} \right.[/tex]

Step 4: Integrate Pt. 3

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:             [tex]\displaystyle \int\limits^{\infty}_0 {14e^{-7x}} \, dx = \lim_{b \to \infty} -2 \int\limits^b_0 {-7e^{-7x}} \, dx[/tex]
  2. [Integral] U-Substitution:                                                                           [tex]\displaystyle \int\limits^{\infty}_0 {14e^{-7x}} \, dx = \lim_{b \to \infty} -2 \int\limits^{-7b}_0 {e^u} \, du[/tex]
  3. [Integral] Apply Exponential Integration:                                                 [tex]\displaystyle \int\limits^{\infty}_0 {14e^{-7x}} \, dx = \lim_{b \to \infty} -2e^u \bigg| \limits^{-7b}_0[/tex]
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:       [tex]\displaystyle \int\limits^{\infty}_0 {14e^{-7x}} \, dx = \lim_{b \to \infty} -2 \Big( e^{-7b} - 1 \Big)[/tex]
  5. Rewrite:                                                                                                     [tex]\displaystyle \int\limits^{\infty}_0 {14e^{-7x}} \, dx = \lim_{b \to \infty} -2 \Big( \frac{1}{e^{7b}} - 1 \Big)[/tex]
  6. Evaluate limit [Limit Rule - Variable Direct Substitution]:                       [tex]\displaystyle \int\limits^{\infty}_0 {14e^{-7x}} \, dx = -2 \Big( \frac{1}{e^{7(\infty)}} - 1 \Big)[/tex]
  7. Simplify:                                                                                                     [tex]\displaystyle \int\limits^{\infty}_0 {14e^{-7x}} \, dx = 2[/tex]

∴ the improper integral is equal to 2 and is also convergent.

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Learn more about improper integrals: https://brainly.com/question/21023287

Learn more about calculus: https://brainly.com/question/23558817

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Topic: AP Calculus BC (Calculus I + II)

Unit: Integration

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