Answer:
[tex]\displaystyle \int\limits^8_1 {(x^2 - 4x + 9)} \, dx = \frac{32452}{5}[/tex]
General Formulas and Concepts:
Calculus
Integration
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]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \int\limits^8_1 {(x^2 - 4x + 9)} \, dx[/tex]
Step 2: Integrate
- [Integral] Rewrite [Integration Property - Addition/Subtraction]: [tex]\displaystyle \int\limits^8_1 {(x^2 - 4x + 9)} \, dx = \int\limits^8_1 {x^2} \, dx - \int\limits^8_1 {4x} \, dx + \int\limits^8_1 {9} \, dx[/tex]
- [Integrals] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int\limits^8_1 {(x^2 - 4x + 9)} \, dx = \int\limits^8_1 {x^2} \, dx - 4\int\limits^8_1 {x} \, dx + 9\int\limits^8_1 {} \, dx[/tex]
- [Integrals] Reverse Power Rule: [tex]\displaystyle \int\limits^8_1 {(x^2 - 4x + 9)} \, dx = \frac{x^3}{3} \bigg| \limits^8_1 - 4 \bigg( \frac{x^2}{2} \bigg) \bigg| \limits^8_1 + 9(x) \bigg| \limits^8_1[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^8_1 {(x^2 - 4x + 9)} \, dx = \frac{511}{3} - 4 \bigg( \frac{63}{2} \bigg) + 9(7)[/tex]
- Simplify: [tex]\displaystyle \int\limits^8_1 {(x^2 - 4x + 9)} \, dx = \frac{32452}{5}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
