Respuesta :

Space

Answer:

[tex]\displaystyle A = \frac{1}{2}[/tex]

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]:                                                                [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:                                     [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Addition/Subtraction]:                                                       [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]  

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

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.

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

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

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

Book: College Calculus 10e

Ver imagen Space