Find the area of the region under the graph of the function f on the interval [1, 8].

f(x) =3/x
square units

Question

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Respuesta :

Space Space · Answer 1 · 2021-09-09T22:50:34+02:00

Answer:

[tex]\displaystyle \int\limits^8_1 {\frac{3}{x}} \, dx = 9 \ln 2[/tex]

General Formulas and Concepts:

Calculus

Integration

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{3}{x} \\\left[ 1 ,\ 8 \right][/tex]

Step 2: Integrate

Substitute in variables [Area of a Region Formula]: [tex]\displaystyle \int\limits^8_1 {\frac{3}{x}} \, dx[/tex]
[Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle 3 \int\limits^8_1 {\frac{1}{x}} \, dx[/tex]
[Integral] Logarithmic Integration: [tex]\displaystyle 3 \int\limits^8_1 {\frac{1}{x}} \, dx = 3 \ln \big| x \big| \bigg| \limits^8_1[/tex]
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle 3 \int\limits^8_1 {\frac{1}{x}} \, dx = 3 \ln 8[/tex]
Simplify: [tex]\displaystyle \int\limits^8_1 {\frac{3}{x}} \, dx = 9 \ln 2[/tex]

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

Unit: Integration