Question

Find the area of the region enclosed by the graphs of the functions

f(x)=x, g(x)=x^3
by partitioning the x- axis.
(Use symbolic notation and fractions where needed.)

Answer 1

`\displaystyle A = (1)/(2)`

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction
8. Left to Right

• Equality Properties

Algebra I

• Terms/Coefficients
• Functions
• Function Notation
• Graphing

Calculus

Area - 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 [Addition/Subtraction]:
`\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx`

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

Explanation:

*Note:

Remember that for the Area of a Region, it is top function minus bottom function.

Also remember that finding area and evaluating are two different things.

Step 1: Define

f(x) = x

g(x) = x³

Bounded (Partitioned) by x-axis

Step 2: Identify Bounds of Integration

Find where the functions intersect (x-values) to determine the bounds of integration.

Simply graph the functions to see where the functions intersect (See Graph Attachment).

Interval: [-1, 1]

1st Integral: [-1, 0]

2nd Integral: [0, 1]

Step 3: Find Area of Region

Integration.

1. Substitute in variables [Area of a Region Formula]:
   `\displaystyle A = \int\limits^0_(-1) {[x^3 - x]} \, dx + \int\limits^1_0 {[x - x^3]} \, dx`
2. [Area] Rewrite Integrals [Integration Property - Subtraction]:
   `\displaystyle A = (\int\limits^0_(-1) {x^3} \, dx - \int\limits^0_(-1) {x} \, dx) + (\int\limits^1_0 {x} \, dx - \int\limits^1_0 {x^3} \, dx)`
3. [Area] [Integrals] Integrate [Integration Rule - Reverse Power Rule]:
   `\displaystyle A = [(x^4)/(4) \bigg|\limits^0_(-1) - ((x^2)/(2)) \bigg|\limits^0_(-1)]+ [(x^2)/(2) \bigg|\limits^1_0 - ((x^4)/(4)) \bigg|\limits^1_0]`
4. [Area] Evaluate [Integration Rule - FTC 1]:
   `\displaystyle A = [(-1)/(4) - ((-1)/(2))] + [(1)/(2) - (1)/(4)]`
5. [Area] [Brackets] Add/Subtract:
   `\displaystyle A = (1)/(4) + (1)/(4)`
6. [Area] Add:
   `\displaystyle A = (1)/(2)`

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

Unit: Area Under the Curve - Area of a Region (Integration)

Book: College Calculus 10e