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Exercise is mixed —some require integration by parts, while others can be integrated by using techniques discussed in the chapter on Integration.
∫^5_0 x √x^2+2 dx.

Respuesta :

Space

Answer:

[tex]\displaystyle \int\limits^5_0 {x\sqrt{x^2 + 2}} \, dx = 27\sqrt{3} - \frac{2\sqrt{2}}{3}[/tex]

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]  

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]

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \int\limits^5_0 {x\sqrt{x^2 + 2}} \, dx[/tex]

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

  1. Set u:                                                                                                             [tex]\displaystyle u = x^2 + 2[/tex]
  2. [u] Basic Power Rule [Derivative Property - Addition/Subtraction]:           [tex]\displaystyle du = 2x \ dx[/tex]
  3. [Limits] Swap:                                                                                                 [tex]\displaystyle \left \{ {{x = 5 ,\ u = 5^2 + 2 = 27} \atop {x = 0 ,\ u = 0^2 + 2 = 2}} \right.[/tex]

Step 3: Integrate Pt. 2

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle \int\limits^5_0 {x\sqrt{x^2 + 2}} \, dx = \frac{1}{2} \int\limits^5_0 {2x\sqrt{x^2 + 2}} \, dx[/tex]
  2. [Integral] U-Substitution:                                                                               [tex]\displaystyle \int\limits^5_0 {x\sqrt{x^2 + 2}} \, dx = \frac{1}{2} \int\limits^{27}_2 {\sqrt{u}} \, du[/tex]
  3. [Integral] Integration Rule [Reverse Power Rule]:                                       [tex]\displaystyle \int\limits^5_0 {x\sqrt{x^2 + 2}} \, dx = \frac{1}{2} \Bigg( \frac{2x^\bigg{\frac{3}{2}}}{3} \Bigg) \Bigg| \limits^{27}_2[/tex]
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:          [tex]\displaystyle \int\limits^5_0 {x\sqrt{x^2 + 2}} \, dx = \frac{1}{2} \bigg( 54\sqrt{3} - \frac{4\sqrt{2}}{3} \bigg)[/tex]
  5. Simplify:                                                                                                         [tex]\displaystyle \int\limits^5_0 {x\sqrt{x^2 + 2}} \, dx = 27\sqrt{3} - \frac{2\sqrt{2}}{3}[/tex]

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

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