Respuesta :
Answer:
[tex]\displaystyle A = 300[/tex]
General Formulas and Concepts:
Calculus
Integrals
- Definite Integrals
- Area under the curve
- Integration Constant C
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
f(x) = 6x + 19
Interval [12, 15]
Step 2: Find Area
- Substitute in variables [Area of a Region Formula]: [tex]\displaystyle A = \int\limits^{15}_{12} {(6x + 19)} \, dx[/tex]
- [Integral] Rewrite [Integration Property - Addition/Subtraction]: [tex]\displaystyle A = \int\limits^{15}_{12} {6x} \, dx + \int\limits^{15}_{12} {19} \, dx[/tex]
- [Integrals] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle A = 6\int\limits^{15}_{12} {x} \, dx + 19\int\limits^{15}_{12} {} \, dx[/tex]
- [Integrals] Integrate [Integration Rule - Reverse Power Rule]: [tex]\displaystyle A = 6(\frac{x^2}{2}) \bigg| \limits^{15}_{12} + 19(x) \bigg| \limits^{15}_{12}[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle A = 6(\frac{81}{2}) + 19(3)[/tex]
- Simplify: [tex]\displaystyle A = 300[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e