Answer:
A. [tex]\displaystyle \int\limits^1_0 {(x^2 + 2x^4)} \, dx[/tex]
General Formulas and Concepts:
Algebra I
Functions
- Function Notation
- Graphing
Calculus
Integration
- Integrals
- Definite Integrals
- Integration Constant C
- Area under the curve/area between 2 curves
Area of a Region Formula: [tex]\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx[/tex]
- f(x) is top function, g(x) is bottom function
Step-by-step explanation:
Step 1: Define
Identify
f(x) = -2x⁴
g(x) = x²
Interval bound [0, 1]
Step 2: Visualize
Graph the given and identify more information. See attachment.
Top function: g(x)
Bottom function: f(x)
Bounds of Integration: [0, 1]
Step 3: Find Area
- Substitute in variables [Area of a Region Formula]: [tex]\displaystyle A = \int\limits^1_0 {[x^2 - (-2x^4)]} \, dx[/tex]
- [Integral] Simplify: [tex]\displaystyle A = \int\limits^1_0 {(x^2 + 2x^4)} \, dx[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e
