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15.10.2019 •

Mathematics

Jordan started at her home at point h she ran to the bank library post office at point b the café and then back to her home the coordinates the position in the miles of each of these locations with respect messenger tell which is located at the origin what is the total distance that jordan ran

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Ответ:

BrainzOli7408

17.05.2022

Mathematics

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

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

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

f(x)=x, g(x)=x^3
by partitionin

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