Answer:
[tex]\displaystyle A = \frac{20\sqrt{15}}{3}[/tex]
General Formulas and Concepts:
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
- Graphing
- Exponential Rule [Root Rewrite]: [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]
Calculus
Derivatives
Derivative Notation
Derivative of a constant is 0
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Area - Integrals
U-Substitution
Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Integration Property [Addition/Subtraction]: [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/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]
Step-by-step explanation:
Step 1: Define
F: y = √(15 - x)
G: y = √(15 - 3x)
H: y = 0
Step 2: Find Bounds of Integration
Solve each equation for the x-value for our bounds of integration.
F
- Set y = 0: 0 = √(15 - x)
- [Equality Property] Square both sides: 0 = 15 - x
- [Subtraction Property of Equality] Isolate x term: -x = -15
- [Division Property of Equality] Isolate x: x = 15
G
- Set y = 0: 0 = √(15 - 3x)
- [Equality Property] Square both sides: 0 = 15 - 3x
- [Subtraction Property of Equality] Isolate x term: -3x = -15
- [Division Property of Equality] Isolate x: x = 5
This tells us that our bounds of integration for function F is from 0 to 15 and our bounds of integration for function G is 0 to 5.
We see that we need to subtract function G from function F to get our area of the region (See attachment graph for visual).
Step 3: Find Area of Region
Integration Part 1
- Rewrite Area of Region Formula [Integration Property - Subtraction]: [tex]\displaystyle A = \int\limits^b_a {f(x)} \, dx - \int\limits^d_c {g(x)} \, dx[/tex]
- [Integral] Substitute in variables and limits [Area of Region Formula]: [tex]\displaystyle A = \int\limits^{15}_0 {\sqrt{15 - x}} \, dx - \int\limits^5_0 {\sqrt{15 - 3x}} \, dx[/tex]
- [Area] [Integral] Rewrite [Exponential Rule - Root Rewrite]: [tex]\displaystyle A = \int\limits^{15}_0 {(15 - x)^{\frac{1}{2}}} \, dx - \int\limits^5_0 {(15 - 3x)^{\frac{1}{2}}} \, dx[/tex]
Step 4: Identify Variables
Set variables for u-substitution for both integrals.
Integral 1:
u = 15 - x
du = -dx
Integral 2:
z = 15 - 3x
dz = -3dx
Step 5: Find Area of Region
Integration Part 2
- [Area] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle A = -\int\limits^{15}_0 {-(15 - x)^{\frac{1}{2}}} \, dx + \frac{1}{3}\int\limits^5_0 {-3(15 - 3x)^{\frac{1}{2}}} \, dx[/tex]
- [Area] U-Substitution: [tex]\displaystyle A = -\int\limits^0_{15} {u^{\frac{1}{2}}} \, du + \frac{1}{3}\int\limits^0_{15} {z^{\frac{1}{2}}} \, dz[/tex]
- [Area] Reverse Power Rule: [tex]\displaystyle A = -(\frac{2u^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15} + \frac{1}{3}(\frac{2z^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15}[/tex]
- [Area] Evaluate [Integration Rule - FTC 1]: [tex]\displaystyle A = -(-10\sqrt{15}) + \frac{1}{3}(-10\sqrt{15})[/tex]
- [Area] Multiply: [tex]\displaystyle A = 10\sqrt{15} + \frac{-10\sqrt{15}}{3}[/tex]
- [Area] Add: [tex]\displaystyle A = \frac{20\sqrt{15}}{3}[/tex]
Topic: AP Calculus AB/BC (Calculus I/II)
Unit: Area Under the Curve - Area of a Region (Integration)
Book: College Calculus 10e
