HELP WITH CALCULUS

Find the area bounded by the curve y = x^2 and the straight line y = 2 + x.
A. 4 1/2
B. 4 1/6
C. 5 1/2
D. 5 1/6

Respuesta :

Space

Answer:

A. 4 1/2

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

Algebra I

Functions

  • Function Notation
  • Graphing

Solving systems of equations by graphing

Calculus

Integration

  • Integrals
  • Definite Integrals
  • Integration Constant C
  • 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

y = x²

y = 2 + x

Step 2: Identify

See Attachment. Find other necessary information.

Interval [-1, 2]

Step 3: Find Area

  1. Substitute in variables [Area of a Region Formula]:                                   [tex]\displaystyle A = \int\limits^2_{-1} {(2 + x - x^2)} \, dx[/tex]
  2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:               [tex]\displaystyle A = \int\limits^2_{-1} {2} \, dx + \int\limits^2_{-1} {x} \, dx - \int\limits^2_{-1} x^2} \, dx[/tex]
  3. [1st Integral] Rewrite [Integration Property - Multiplied Constant]:             [tex]\displaystyle A = 2\int\limits^2_{-1} {} \, dx + \int\limits^2_{-1} {x} \, dx - \int\limits^2_{-1} x^2} \, dx[/tex]
  4. [Integrals] Integrate [Integration Rule - Reverse Power Rule]:                   [tex]\displaystyle A = 2(x) \bigg| \limits^2_{-1} + \frac{x^2}{2} \bigg| \limits^2_{-1} - \frac{x^3}{3} \bigg| \limits^2_{-1}[/tex]
  5. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           [tex]\displaystyle A = 2(3) + \frac{3}{2} - 3[/tex]
  6. Simplify:                                                                                                         [tex]\displaystyle A = \frac{9}{2}[/tex]
  7. Reduce:                                                                                                         [tex]\displaystyle A = 4\frac{1}{2}[/tex]

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

Unit: Integration

Book: College Calculus 10e

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