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Mathematics, 17.07.2019 13:00 tcarston1728

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

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Answer from: rocksquad7917

A. 4 1/2

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Algebra I

Functions

Function NotationGraphing

Solving systems of equations by graphing

Calculus

Integration

IntegralsDefinite IntegralsIntegration Constant CArea under the curve

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

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

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

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

Unit: Integration

Book: College Calculus 10e

With calculus find the area bounded by the curve y = x^2 and the straight line y = 2 + x. a. 4 1/2

ansver

Answer from: Quest

4a. is 3 miles every hour

4b. is 1/3 of an hour per mile

step-by-step explanation:

Middle school. ratios.rita walks. for me

ansver

Answer from: Quest

correct choice is a

step-by-step explanation:

the equation of the line passing through the points $(x_1,y_1)$ and $(x_2,y_2)$ is

$\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}.$

thus, the equation of the line that passes through the points (0, -6) and (-3, 0) is

$\dfrac{x-0}{-3-0}=\dfrac{)}{)},\\ \\\dfrac{x}{-3}=\dfrac{y+6}{6},\\ \\2x=-y-6,\\ \\2x+y+6=0,\\ \\y=-2x-6.$

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