Respuesta :

Answer:

[tex]\int^2_1 ln 5x dx=1.99[/tex]

Step-by-step explanation:

we have the integral:

[tex]\int^2_1 ln 5x dx[/tex]

The formula to integrate by parts is:

[tex]\int udv=uv-\int vdu[/tex]

in this case we will use [tex]u=ln5x[/tex] and [tex]dv=dx[/tex].

so [tex]v=x[/tex]

and using logarithm derivation [tex]d(lnw)=\frac{dw}{w}[/tex]

[tex]du=\frac{5}{5x} =\frac{1}{x}[/tex]

thus:

[tex]\int ln 5x dx=(ln5x)(x)- \int x(\frac{1}{x})dx[/tex]

[tex]\int ln 5x dx=xln5x - \int dx[/tex]

[tex]\int ln 5x dx=xln5x - x[/tex]

evaluating the limits:

[tex]\int^2_1 ln 5x dx=[xln5x - x]^2_1\\=[2ln(5*2)- 2]-[1ln(5*1) - 1]\\=[2ln10-2]-[ln5-1]\\=[2(2.3)-2]-[1.61-1]\\=[4.6-2]-0.61\\=2.6-0.61\\=1.99[/tex]

Space

Answer:

[tex]\displaystyle \int\limits^2_1 {\ln 5x} \, dx = \ln(20) - 1[/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]

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

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 by Parts:                                                                                               [tex]\displaystyle \int {u} \, dv = uv - \int {v} \, du[/tex]

  • [IBP] LIPET: Logs, inverses, Polynomials, Exponentials, Trig

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \int\limits^2_1 {\ln 5x} \, dx[/tex]

Step 2: Integrate Pt. 1

Identify variables for integration by parts using LIPET.

  1. Set u:                                                                                                             [tex]\displaystyle u = \ln 5x[/tex]
  2. [u] Logarithmic Differentiation [Derivative Rule - Chain Rule]:                  [tex]\displaystyle du = \frac{(5x)'}{5x} \ dx[/tex]
  3. [du] Basic Power Rule [Derivative Rule - Multiplied Constant]:                 [tex]\displaystyle du = \frac{5}{5x} \ dx[/tex]
  4. [du] Simplify:                                                                                                   [tex]\displaystyle du = \frac{1}{x} \ dx[/tex]
  5. Set dv:                                                                                                           [tex]\displaystyle dv = dx[/tex]
  6. [dv] Integration Rule [Reverse Power Rule]:                                               [tex]\displaystyle v = x[/tex]

Step 3: Integrate Pt. 2

  1. [Integral] Integration by Parts:                                                                       [tex]\displaystyle \int\limits^2_1 {\ln 5x} \, dx = x \ln(5x) \bigg| \limits^2_1 - \int\limits^2_1 {} \, dx[/tex]
  2. [Integral] Integration Rule [Reverse Power Rule]:                                       [tex]\displaystyle \int\limits^2_1 {\ln 5x} \, dx = x \ln(5x) \bigg| \limits^2_1 - x \bigg| \limits^2_1[/tex]
  3. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           [tex]\displaystyle \int\limits^2_1 {\ln 5x} \, dx = \ln(20) - 1[/tex]

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

Unit: Integration

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