Question

Find the area of the region enclosed by the graphs of these equations. (CALCULUS HELP)

Answer 1

Answer:

$\displaystyle A = \frac{20\sqrt{15}}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Equality Properties

1. Multiplication Property of Equality
2. Division Property of Equality
3. Addition Property of Equality
4. Subtraction Property of Equality

Algebra I

• Terms/Coefficients
• Graphing
• Exponential Rule [Root Rewrite]:                                                                   $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

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

Integration Property [Multiplied Constant]:                                                         $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

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

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

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

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

1. Set y = 0:                                                                                                         0 = √(15 - x)
2. [Equality Property] Square both sides:                                                          0 = 15 - x
3. [Subtraction Property of Equality] Isolate x term:                                         -x = -15
4. [Division Property of Equality] Isolate x:                                                        x = 15

G

1. Set y = 0:                                                                                                         0 = √(15 - 3x)
2. [Equality Property] Square both sides:                                                          0 = 15 - 3x
3. [Subtraction Property of Equality] Isolate x term:                                         -3x = -15
4. [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

1. Rewrite Area of Region Formula [Integration Property - Subtraction]:     $\displaystyle A = \int\limits^b_a {f(x)} \, dx - \int\limits^d_c {g(x)} \, dx$
2. [Integral] Substitute in variables and limits [Area of Region Formula]:     $\displaystyle A = \int\limits^{15}_0 {\sqrt{15 - x}} \, dx - \int\limits^5_0 {\sqrt{15 - 3x}} \, dx$
3. [Area] [Integral] Rewrite [Exponential Rule - Root Rewrite]:                       $\displaystyle A = \int\limits^{15}_0 {(15 - x)^{\frac{1}{2}}} \, dx - \int\limits^5_0 {(15 - 3x)^{\frac{1}{2}}} \, dx$

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

1. [Area] Rewrite [Integration Property - Multiplied Constant]:                       $\displaystyle A = -\int\limits^{15}_0 {-(15 - x)^{\frac{1}{2}}} \, dx + \frac{1}{3}\int\limits^5_0 {-3(15 - 3x)^{\frac{1}{2}}} \, dx$
2. [Area] U-Substitution:                                                                                   $\displaystyle A = -\int\limits^0_{15} {u^{\frac{1}{2}}} \, du + \frac{1}{3}\int\limits^0_{15} {z^{\frac{1}{2}}} \, dz$
3. [Area] Reverse Power Rule:                                                                         $\displaystyle A = -(\frac{2u^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15} + \frac{1}{3}(\frac{2z^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15}$
4. [Area] Evaluate [Integration Rule - FTC 1]:                                                   $\displaystyle A = -(-10\sqrt{15}) + \frac{1}{3}(-10\sqrt{15})$
5. [Area] Multiply:                                                                                               $\displaystyle A = 10\sqrt{15} + \frac{-10\sqrt{15}}{3}$
6. [Area] Add:                                                                                                     $\displaystyle A = \frac{20\sqrt{15}}{3}$

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

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

Book: College Calculus 10e