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

Answer:

$\displaystyle A = 2b^3$

General Formulas and Concepts:

Calculus

Integration

• Integrals

Integration Rule [Reverse Power Rule]:                                                               $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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:                                                                                     $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step 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]:                     $\displaystyle A = 6 \bigg( \frac{x^3}{3} \bigg) \bigg| \limits^b_0$
4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           $\displaystyle A = 6 \bigg( \frac{b^3}{3} \bigg)$
5. Simplify:                                                                                                         $\displaystyle A = 2b^3$

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

Unit: Integration