Find the area of the region under the graph of the function f on the interval [4, 8].
f(x) = 10/x^2

If you could show the steps and the answer, I would really appreciate it!

Respuesta :

Space

Answer:

[tex]\displaystyle \int\limits^8_4 {\frac{10}{x^2}} \, dx = \frac{5}{4}[/tex]

General Formulas and Concepts:

Calculus

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]

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 f(x) = \frac{10}{x^2} \\\left[ 4 ,\ 8 \right][/tex]

Step 2: Find Area

  1. Substitute in variables [Area of a Region Formula]:                                   [tex]\displaystyle \int\limits^8_4 {\frac{10}{x^2}} \, dx[/tex]
  2. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle \int\limits^8_4 {\frac{10}{x^2}} \, dx = 10 \int\limits^8_4 {\frac{1}{x^2}} \, dx[/tex]
  3. [Integral] Integrate [Integration Rule - Reverse Power Rule]:                     [tex]\displaystyle \int\limits^8_4 {\frac{10}{x^2}} \, dx = 10 \bigg( \frac{-1}{x} \bigg) \bigg| \limits^8_4[/tex]
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           [tex]\displaystyle \int\limits^8_4 {\frac{10}{x^2}} \, dx = 10 \bigg( \frac{1}{8} \bigg)[/tex]
  5. Simplify:                                                                                                        [tex]\displaystyle \int\limits^8_4 {\frac{10}{x^2}} \, dx = \frac{5}{4}[/tex]

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

Unit: Integration

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