contestada

Evaluate the definite integral (S is the integration symbols with the indicated values)
1
S 4x(6Sqrt(1+x^2) dx
0

Respuesta :

Space

Answer:

[tex]\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 8(2\sqrt{2} - 1)[/tex]

General Formulas and Concepts:

Algebra I

Terms/Coefficients

  • Expanding/Factoring

Exponential Property [Root Rewrite]:                                                                   [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]

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^1_0 {4x(6\sqrt{1 + x^2})} \, dx[/tex]

Step 2: Integrate Pt. 1

  1. [Integrand] Simplify:                                                                                       [tex]\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = \int\limits^1_0 {24x\sqrt{1 + x^2}} \, dx[/tex]
  2. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 24\int\limits^1_0 {x\sqrt{1 + x^2}} \, dx[/tex]

Step 3: integrate Pt. 2

Identify variables for u-substitution.

  1. Set u:                                                                                                             [tex]\displaystyle u = 1 + x^2[/tex]
  2. [u] Differentiate [Basic Power Rule, Derivative Properties]:                       [tex]\displaystyle du = 2x \ dx[/tex]
  3. [Bounds] Switch:                                                                                           [tex]\displaystyle \left \{ {{x = 1 ,\ u = 1 + 1^2 = 2} \atop {x = 0 ,\ u = 1 + 0^2 = 1}} \right.[/tex]

Step 4: Integrate Pt. 3

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 12\int\limits^1_0 {2x\sqrt{1 + x^2}} \, dx[/tex]
  2. [Integral] U-Substitution:                                                                               [tex]\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 12\int\limits^2_1 {\sqrt{u}} \, du[/tex]
  3. [Integrand] Rewrite [Exponential Property - Root Rewrite]:                       [tex]\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 12\int\limits^2_1 {u^\big{\frac{1}{2}}} \, du[/tex]
  4. [Integral] Reverse Power Rule:                                                                     [tex]\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 12 \bigg( \frac{2u^\big{\frac{3}{2}}}{3} \bigg) \bigg| \limits^2_1[/tex]
  5. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           [tex]\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 12 \bigg( \frac{4\sqrt{2}}{3} - \frac{2}{3} \bigg)[/tex]
  6. Factor:                                                                                                           [tex]\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 8(2\sqrt{2} - 1)[/tex]

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

Unit: Integration

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