Respuesta :

Space

Answer:

[tex]\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy = \frac{20}{3}[/tex]

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

Calculus

Integrals

  • Definite Integrals

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]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy[/tex]

Step 2: Integrate

  1. [Integral] Rewrite [Integration Property - Addition/Subtraction]:               [tex]\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy = \int\limits^4_0 {} \, dy + \int\limits^4_0 {3y} \, dy - \int\limits^4_0 {y^2} \, dy[/tex]
  2. [2nd Integral] Rewrite [Integration Property - Multiplied Constant]:         [tex]\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy = \int\limits^4_0 {} \, dy + 3\int\limits^4_0 {y} \, dy - \int\limits^4_0 {y^2} \, dy[/tex]
  3. [Integrals] Reverse Power Rule:                                                                   [tex]\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy = y \bigg| \limits^4_0 + 3(\frac{y^2}{2}) \bigg| \limits^4_0 - (\frac{y^3}{3}) \bigg| \limits^4_0[/tex]
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           [tex]\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy = 4 + 3(8) - \frac{64}{3}[/tex]
  5. Simplify:                                                                                                         [tex]\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy = \frac{20}{3}[/tex]

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

Unit: Integration

Book: College Calculus 10e