Question

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

Answer 1

Answer:

$\displaystyle A = \frac{8}{21}$

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

• Multiplication Property of Equality
• Division Property of Equality
• Addition Property of Equality
• Subtraction Property of Equality

Algebra I

• Terms/Coefficients
• Functions
• Function Notation
• Graphing
• Solving systems of equations

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.

Step 1: Define

f(x) = x²

g(x) = x⁶

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]

Lower bound: -1

Upper Bound: 1

Step 3: Find Area of Region

Integration

1. Substitute in variables [Area of a Region Formula]:                                     $\displaystyle A = \int\limits^1_{-1} {[x^2 - x^6]} \, dx$
2. [Area] Rewrite [Integration Property - Subtraction]:                                     $\displaystyle A = \int\limits^1_{-1} {x^2} \, dx - \int\limits^1_{-1} {x^6} \, dx$
3. [Area] Integrate [Integration Rule - Reverse Power Rule]:                           $\displaystyle A = \frac{x^3}{3} \bigg| \limit^1_{-1} - \frac{x^7}{7} \bigg| \limit^1_{-1}$
4. [Area] Evaluate [Integration Rule - FTC 1]:                                                    $\displaystyle A = \frac{2}{3} - \frac{2}{7}$
5. [Area] Subtract:                                                                                               $\displaystyle A = \frac{8}{21}$

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

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

Book: College Calculus 10e