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How would you use the Fundamental Theorem of Calculus to determine the value(s) of b if the area under the graph g(x)=4x between x=1 and x=b is equal to 240?

Respuesta :

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Answer:

[tex]\displaystyle b = 11[/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  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

Algebra I

Functions

  • Function Notation

Calculus

Integration

  • 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]

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

g(x) = 4x

Interval [1, b]

A = 240

Step 2: Solve for b

  1. Substitute in variables [Area of a Region Formula]:                                   [tex]\displaystyle \int\limits^b_1 {4x} \, dx = 240[/tex]
  2. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle 4\int\limits^b_1 {x} \, dx = 240[/tex]
  3. [Integral] Integrate [Integration Rule - Reverse Power Rule]:                     [tex]\displaystyle 4(\frac{x^2}{2}) \bigg| \limits^b_1 = 240[/tex]
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           [tex]\displaystyle 4(\frac{b^2}{2} - \frac{1}{2}) = 240[/tex]
  5. [Distributive Property] Distribute 4:                                                             [tex]\displaystyle 2b^2 - 2 = 240[/tex]
  6. [Addition Property of Equality] Add 2 on both sides:                                 [tex]\displaystyle 2b^2 = 242[/tex]
  7. [Division Property of Equality] Divide 2 on both sides:                               [tex]\displaystyle b^2 = 121[/tex]
  8. [Equality Property] Square root both sides:                                                 [tex]\displaystyle b = \pm 11[/tex]
  9. Choose:                                                                                                         [tex]\displaystyle b = 11[/tex]

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

Unit: Integration

Book: College Calculus 10e

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