Answer:
[tex]\displaystyle A = \frac{12}{5}[/tex]
General Formulas and Concepts:
Math
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
Algebra I
- Terms/Coefficients
- Factoring
- Coordinates (x, y)
- Solving systems of equations using substitution/elimination
- Solving systems of equations by graphing
- Function Notation
- Interval Notation
Calculus
Integration
Integration Property [Addition/Subtraction]: [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Integration Property [Splitting Integral]: [tex]\displaystyle \int\limits^c_a {f(x)} \, dx = \int\limits^b_a {f(x)} \, dx + \int\limits^c_b {f(x)} \, dx[/tex]
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]
Area of a Region Formula: [tex]\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx[/tex]
- f(x) is always top function
- g(x) is always bottom function
- "Top minus Bottom"
Step-by-step explanation:
Step 1: Define
Identify bounded region. See attached graph.
y = 3x - 6
y = -2x + 8
Bounded by x-axis and between those 2 lines (already pre-determined; taking an integral always takes it to the x-axis).
Step 2: Analyze Graph
See attached graph.
Looking at our systems of equations on the graph, we see that our limits of integration is from x = 2 to x = 4.
We don't have a continuous top function through the interval [2, 4] (it switches from y = 3x - 6 to y = -2x + 8), so we need to split it into 2 integrals to find the total area.
We can either use the graph and identify the intersection point, which is x = 2.8, or we can solve it algebraically (systems of equations - substitution method):
- Substitute in y: 3x - 6 = -2x + 8
- [Addition Property of Equality] Add 2x on both sides: 5x - 6 = 8
- [Addition Property of Equality] Add 6 on both sides: 5x = 14
- [Division Property of Equality] Divide 5 on both sides: x = 14/5
Our 2 intervals would be [2, 14/5] and [14/5, 4] for their respective integrals.
Step 3: Find Area
Our top functions are the linear lines y = 3x - 6 and y = -2x + 8 and our continuous bottom function is the x-axis (x = 0).
We can redefine the linear lines as f₁(x) = 3x - 6, f₂(x) = -2x + 8, and g(x) = 0.
Integration
- [Area of a Region Formula] Rewrite/Redefine [Int Prop SI]: [tex]\displaystyle A = \int\limits^b_a {[f_1(x) - g(x)]} \, dx + \int\limits^c_b {[f_2(x) - g(x)]} \, dx[/tex]
- [Area of a Region Formula] Substitute in variables: [tex]\displaystyle A = \int\limits^{\frac{14}{5}}_2 {[(3x - 6) - 0]} \, dx + \int\limits^4_{\frac{14}{5}} {[(-2x + 8) - 0]} \, dx[/tex]
- [Integrals] Simplify Integrands: [tex]\displaystyle A = \int\limits^{\frac{14}{5}}_2 {[3x - 6]} \, dx + \int\limits^4_{\frac{14}{5}} {[-2x + 8]} \, dx[/tex]
- [Integrals - Algebra] Factor: [tex]\displaystyle A = \int\limits^{\frac{14}{5}}_2 {[3(x - 2)]} \, dx + \int\limits^4_{\frac{14}{5}} {[-2(x - 4)]} \, dx[/tex]
- [Integrals] Simplify [Int Prop MC]: [tex]\displaystyle A = 3 \int\limits^{\frac{14}{5}}_2 {[x - 2]} \, dx - 2 \int\limits^4_{\frac{14}{5}} {[x - 4]} \, dx[/tex]
- [Integrals] Integrate [Int Rule RPR]: [tex]\displaystyle A = 3(\frac{x^2}{2} - 2x) \bigg| \limits^{\frac{14}{5}}_2 - 2(\frac{x^2}{2} - 4x) \bigg| \limits^4_{\frac{14}{5}}[/tex]
- [Integrals] Evaluate [Int Rule FTC 1]: [tex]\displaystyle A = 3(\frac{8}{25}) - 2(\frac{-18}{25})[/tex]
- [Expression] Multiply: [tex]\displaystyle A = \frac{24}{25} + \frac{36}{25}[/tex]
- [Expression] Add: [tex]\displaystyle A = \frac{12}{5}[/tex]
We have found the area bounded by the x-axis and linear lines y = 3x - 6 and y = -2x + 8.
Topic: Calculus BC
Unit: Area between 2 Curves, Volume, Arc Length, Surface Area
Chapter 7 (College Textbook - Calculus 10e)
Hope this helped!
