Question

Use a definite integral to find an expression that represents the area of the region between the given curve and the x-axis on the interval [0,b].

y=6x^2

Answer 1

`\displaystyle A = 2b^3`

General Formulas and Concepts:

Calculus

Integration

• Integrals

Integration Rule [Reverse Power Rule]:
`\displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + C`

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

y = 6x²

[0, b]

Step 2: Find Area

1. Substitute in variables [Area of a Region Formula]:
   `\displaystyle A = \int\limits^b_0 {6x^2} \, dx`
2. [Integral] Rewrite [integration Property - Multiplied Constant]:
   `\displaystyle A = 6\int\limits^b_0 {x^2} \, dx`
3. [Integral] integrate [Integration Rule - Reverse Power Rule]:
   `\displaystyle A = 6 \bigg( (x^3)/(3) \bigg) \bigg| \limits^b_0`
4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
   `\displaystyle A = 6 \bigg( (b^3)/(3) \bigg)`
5. Simplify:
   `\displaystyle A = 2b^3`

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

Unit: Integration