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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.
∫^1_0 x^2 dx/2x^3+1

Respuesta :

Space

Answer:

[tex]\displaystyle \int\limits^1_0 {\frac{x^2}{2x^3 + 1}} \, dx = \frac{\ln 3}{6}[/tex]

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

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]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \int\limits^1_0 {\frac{x^2}{2x^3 + 1}} \, dx[/tex]

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

  1. Set u:                                                                                                             [tex]\displaystyle u = 2x^3 + 1[/tex]
  2. [u] Basic Power Rule [Derivative Properties]:                                             [tex]\displaystyle du = 6x^2 \ dx[/tex]
  3. [Limits] Swap:                                                                                                 [tex]\displaystyle \left \{ {{x = 1 ,\ u = 2(1)^3 + 1 = 3} \atop {x = 0 ,\ u = 2(0)^3 + 1 = 1}} \right.[/tex]

Step 3: Integrate Pt. 2

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle \int\limits^1_0 {\frac{x^2}{2x^3 + 1}} \, dx = \frac{1}{6} \int\limits^1_0 {\frac{6x^2}{2x^3 + 1}} \, dx[/tex]
  2. [Integral] U-Substitution:                                                                               [tex]\displaystyle \int\limits^1_0 {\frac{x^2}{2x^3 + 1}} \, dx = \frac{1}{6} \int\limits^3_1 {\frac{1}{u}} \, du[/tex]
  3. [Integral] Logarithmic Integration:                                                               [tex]\displaystyle \int\limits^1_0 {\frac{x^2}{2x^3 + 1}} \, dx = \frac{1}{6} \ln \big| u \big| \bigg| \limits^3_1[/tex]
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           [tex]\displaystyle \int\limits^1_0 {\frac{x^2}{2x^3 + 1}} \, dx = \frac{\ln 3}{6}[/tex]

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

Unit: Integration

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