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Given the hyperbola y = 1/x, find the area under the curve between x = 5 and x = 33 to 3 sig. dig. namely: integral subscript 5 superscript 33 1 over x d x space equals . State the definite integral and evaluate it:Given the hyperbola y = 1/x, find the area under the curve between x = 5 and x = 33 to 3 sig. dig. namely: integral subscript 5 superscript 33 1 over x d x space equals . State the definite integral and evaluate it:

Respuesta :

Space

Answer:

[tex]\displaystyle A = \int\limits^{33}_{5} {\frac{1}{x}} \, dx[/tex]

[tex]\displaystyle A = \ln \frac{33}{5}[/tex]

General Formulas and Concepts:

Algebra II

  • Logarithmic Properties

Calculus

Integration

  • Integrals

Integration Rule [Fundamental Theorem of Calculus 1]:                                     [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Area of a Region Formula:                                                                                     [tex]\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx[/tex]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle y = \frac{1}{x}[/tex]

Bounds: [5, 33]

Step 2: Find Area

  1. Substitute in variables [Area of a Region Formula]:                                   [tex]\displaystyle A = \int\limits^{33}_{5} {\frac{1}{x}} \, dx[/tex]
  2. [Integral] Integrate [Logarithmic Integration]:                                             [tex]\displaystyle A = \ln |x| \bigg| \limits^{33}_5[/tex]
  3. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           [tex]\displaystyle A = \ln |33| - \ln |5|[/tex]
  4. Condense:                                                                                                     [tex]\displaystyle A = \ln \frac{33}{5}[/tex]

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

Unit: Integration

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