Mathematics, 22.03.2021 18:50 nayelieangueira

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.)

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Answer from: bellbug08

$\displaystyle A = \frac{1}{2}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to RightEquality Properties

Algebra I

Terms/CoefficientsFunctionsFunction NotationGraphing

Calculus

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

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

Step-by-step 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.

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

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

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

Book: College Calculus 10e