Respuesta :

Space

Answer:

The improper integral diverges.

[tex]\displaystyle \int\limits^{\infty}_0 {8e^{-x} - 8x} \, 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]:

  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 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^{\infty}_0 {8e^{-x} - 8x} \, dx[/tex]

Step 2: Integrate Pt. 1

  1. [Integral] Rewrite [Improper Integral]:                                                         [tex]\displaystyle \int\limits^{\infty}_0 {8e^{-x} - 8x} \, dx = \lim_{b \to \infty} \int\limits^{b}_0 {8e^{-x} - 8x} \, dx[/tex]
  2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:              [tex]\displaystyle \int\limits^{\infty}_0 {8e^{-x} - 8x} \, dx = \lim_{b \to \infty} \Bigg[ \int\limits^{b}_0 {8e^{-x}} \, dx - \int\limits^{b}_0 {8x} \, dx \Bigg][/tex]
  3. [Integrals] Rewrite [Integration Property - Multiplied Constant]:              [tex]\displaystyle \int\limits^{\infty}_0 {8e^x - 8x} \, dx = \lim_{b \to \infty} \Bigg[ 8 \int\limits^{b}_0 {e^{-x}} \, dx - 8 \int\limits^{b}_0 {x} \, dx \Bigg][/tex]

Step 3: Integrate Pt. 2

Set variables for u-substitution.

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

Step 4: Integrate Pt. 3

  1. [Left Integral] Rewrite [Integration Property - Multiplied Constant]:         [tex]\displaystyle \int\limits^{\infty}_0 {8e^x - 8x} \, dx = \lim_{b \to \infty} \Bigg[ -8 \int\limits^{b}_0 {-e^{-x}} \, dx - 8 \int\limits^{b}_0 {x} \, dx \Bigg][/tex]
  2. [Left Integral] Apply Integration Method [U-Substitution]:                        [tex]\displaystyle \int\limits^{\infty}_0 {8e^x - 8x} \, dx = \lim_{b \to \infty} \Bigg[ -8 \int\limits^{-b}_0 {e^{u}} \, du - 8 \int\limits^{b}_0 {x} \, dx \Bigg][/tex]
  3. [Left Integral] Apply Exponential Integration:                                            [tex]\displaystyle \int\limits^{\infty}_0 {8e^x - 8x} \, dx = \lim_{b \to \infty} \Bigg[ -8e^u \bigg| \limits^{-b}_0 - 8 \int\limits^{b}_0 {x} \, dx \Bigg][/tex]
  4. [Integral] Apply Integration Rule [Reverse Power Rule]:                           [tex]\displaystyle \int\limits^{\infty}_0 {8e^x - 8x} \, dx = \lim_{b \to \infty} \Bigg[ -8e^u \bigg| \limits^{-b}_0 - 8 \bigg( \frac{x^2}{2} \bigg) \bigg| \limits^{b}_0 \Bigg][/tex]
  5. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:         [tex]\displaystyle \int\limits^{\infty}_0 {8e^x - 8x} \, dx = \lim_{b \to \infty} \Bigg[ -8 \big( e^{-b} - 1 \big) - 8 \bigg( \frac{b^2}{2} \bigg) \Bigg][/tex]
  6. Simplify:                                                                                                        [tex]\displaystyle \int\limits^{\infty}_0 {8e^x - 8x} \, dx = \lim_{b \to \infty} -4 \Bigg[ 2 \big( e^{-b} - 1 \big) - b^2 \Bigg][/tex]
  7. [Limit] Evaluate [Limit Rule - Variable Direct Substitution]:                         [tex]\displaystyle \int\limits^{\infty}_0 {8e^x - 8x} \, dx = -4 \Bigg[ 2 \big( e^{-\infty} - 1 \big) - \infty^2 \Bigg][/tex]
  8. Simplify:                                                                                                        [tex]\displaystyle \int\limits^{\infty}_0 {8e^{-x} - 8x} \, dx = -\infty[/tex]

∴ the improper integral tends to ∞ and is divergent.

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Learn more about improper integrals: https://brainly.com/question/14412090

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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