Question

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

f(x) =3/x
square units

Answer 1

`\displaystyle \int\limits^8_1 {(3)/(x)} \, dx = 9 \ln 2`

General Formulas and Concepts:

Calculus

Integration

• Integrals

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

Integration Property [Multiplied Constant]:
`\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx`

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

Explanation:

Step 1: Define

Identify

`\displaystyle f(x) = (3)/(x) \\\left[ 1 ,\ 8 \right]`

Step 2: Integrate

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

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

Unit: Integration