contestada

Determine whether the improper integral converges or diverges, and find the value of each that converges.
∫^[infinity]_-[infinity] e^|x| dx

Respuesta :

Space

Answer:

The improper integral diverges.

[tex]\displaystyle \int\limits^{\infty}_{- \infty} {e^{|x|}} \, 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 Property [Splitting Integral]:                                                           [tex]\displaystyle \int\limits^c_a {f(x)} \, dx = \int\limits^b_a {f(x)} \, dx + \int\limits^c_b {f(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}_{- \infty} {e^{|x|}} \, dx[/tex]

Step 2: Integrate Pt. 1

  1. [Integral] Rewrite [Integration Property - Splitting Integral]:                   [tex]\displaystyle \int\limits^{\infty}_{- \infty} {e^{|x|}} \, dx = \int\limits^{0}_{- \infty} {e^{|x|}} \, dx + \int\limits^{\infty}_{0} {e^{|x|}} \, dx[/tex]
  2. [Integrals] Rewrite [Improper Integrals]:                                                   [tex]\displaystyle \int\limits^{\infty}_{- \infty} {e^{|x|}} \, dx = \lim_{a \to -\infty} \int\limits^{0}_{a} {e^{|x|}} \, dx + \lim_{b \to \infty} \int\limits^{b}_{0} {e^{|x|}} \, dx[/tex]

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

  1. Set u:                                                                                                         [tex]\displaystyle u = |x|[/tex]
  2. [u] Differentiate [Absolute Value Differentiation]:                                   [tex]\displaystyle du = \frac{|x|}{x} \ dx[/tex]

Step 4: Integrate Pt. 3

  1. [Integrals] Rewrite [Integration Property - Multiplied Constant]:           [tex]\displaystyle \int\limits^{\infty}_{- \infty} {e^{|x|}} \, dx = \lim_{a \to -\infty} \frac{x}{|x|} \int\limits^{0}_{a} {\frac{|x|}{x} e^{|x|}} \, dx + \lim_{b \to \infty} \frac{x}{|x|} \int\limits^{b}_{0} {\frac{|x|}{x} e^{|x|}} \, dx[/tex]
  2. [Integrals] Apply U-Substitution:                                                               [tex]\displaystyle \int\limits^{\infty}_{- \infty} {e^{|x|}} \, dx = \lim_{a \to -\infty} \frac{x}{|x|} \int\limits^{0}_{a} {e^{u}} \, du + \lim_{b \to \infty} \frac{x}{|x|} \int\limits^{b}_{0} {e^{u}} \, du[/tex]
  3. [Integrals] Apply Exponential Integration:                                               [tex]\displaystyle \int\limits^{\infty}_{- \infty} {e^{|x|}} \, dx = \lim_{a \to -\infty} \frac{x}{|x|} \bigg( e^u \bigg) \bigg| \limits^{x = 0}_{x = a} + \lim_{b \to \infty} \frac{x}{|x|} \bigg( e^u \bigg) \bigg| \limits^{x = b}_{x = 0}[/tex]
  4. [u] Back-substitute:                                                                                   [tex]\displaystyle \int\limits^{\infty}_{- \infty} {e^{|x|}} \, dx = \lim_{a \to -\infty} \frac{x}{|x|} \bigg( e^{|x|} \bigg) \bigg| \limits^{0}_{a} + \lim_{b \to \infty} \frac{x}{|x|} \bigg( e^{|x|} \bigg) \bigg| \limits^{b}_{0}[/tex]
  5. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:       [tex]\displaystyle \int\limits^{\infty}_{- \infty} {e^{|x|}} \, dx = \lim_{a \to -\infty} \frac{x \big( 1 - e^{|a|} \big) }{|x|} + \lim_{b \to \infty} \frac{x \big( e^{|b|} - 1 \big)}{|x|}[/tex]
  6. [Limits] Evaluate [Limit Rule - Variable Direct Substitution]:                   [tex]\displaystyle \int\limits^{\infty}_{- \infty} {e^{|x|}} \, dx = \frac{x \big( 1 - e^{\infty} \big) }{|x|} + \frac{x \big( e^{\infty} - 1 \big)}{|x|}[/tex]
  7. Simplify:                                                                                                     [tex]\displaystyle \int\limits^{\infty}_{- \infty} {e^{|x|}} \, dx = \infty[/tex]

∴ the improper integrals tends to ∞ and is divergent.

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Learn more about improper integrals: https://brainly.com/question/14413973

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

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Topic: AP Calculus BC (Calculus I + II)

Unit: Integration

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