Answer:
[tex]\displaystyle A = \frac{32}{3}[/tex]
General Formulas and Concepts:
Calculus
Integrals
- Definite Integrals
- Area under the curve
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]
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]
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
Identify
Curve: x³ + 2x² - 3x
Interval: [-3, 1]
Step 2: Find Area
- Set up: [tex]\displaystyle A = \int\limits^1_{-3} {(x^3 + 2x^2 - 3x)} \, dx[/tex]
- [Integral] Rewrite [Integration Property - Addition/Subtraction]: [tex]\displaystyle A = \int\limits^1_{-3} {x^3} \, dx + \int\limits^1_{-3} {2x^2} \, dx - \int\limits^1_{-3} {3x} \, dx[/tex]
- [Integrals] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle A = \int\limits^1_{-3} {x^3} \, dx + 2\int\limits^1_{-3} {x^2} \, dx - 3\int\limits^1_{-3} {x} \, dx[/tex]
- [Integrals] Integrate [Integration Rule - Reverse Power Rule]: [tex]\displaystyle A = (\frac{x^4}{4}) \bigg| \limits^1_{-3} + 2(\frac{x^3}{3}) \bigg| \limits^1_{-3} - 3(\frac{x^2}{2}) \bigg| \limits^1_{-3}[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle A = -20 + 2(\frac{28}{3}) - 3(-4)[/tex]
- Evaluate: [tex]\displaystyle A = \frac{32}{3}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e
