Mathematics, 29.05.2021 01:30 meganwintergirl

Consider the following functions. (See attachment)
Find the area of the region.

Answers: 2

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Answer from: hbked23

Area: $\displaystyle \frac{1}{2}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Algebra I

FunctionsFunction NotationGraphing FunctionsExponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Area - Integrals

Area under a curveArea between 2 curves

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

U-Substitution

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

Step-by-step explanation:

Step 1: Define

Identify functions

$\displaystyle f(x) = \sqrt[3]{x - 9}$

$\displaystyle g(x) = x - 9$

Step 2: Identify Info

Graph the functions - See Attachment

[1st Integral] Bounds: [8, 9], g(x) top function/f(x) bottom function

[2nd Integral] Bounds: [9, 10], f(x) top function/g(x) bottom function

Step 3: Find Area Pt. 1

Set up integrals [Area of a Region Formula]: $\displaystyle A = \int\limits^9_8 {[(x - 9) - \sqrt[3]{x - 9}]} \, dx + \int\limits^{10}_9 {[\sqrt[3]{x - 9} - (x - 9)]} \, dx$ Rewrite integrals [Integration Property - Addition/Subtraction]: $\displaystyle A = \int\limits^9_8 {(x - 9)} \, dx - \int\limits^9_8 {\sqrt[3]{x - 9}} \, dx + \int\limits^{10}_9 {\sqrt[3]{x - 9}} \, dx - \int\limits^{10}_9 {(x - 9)} \, dx$ Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = (x^2 - 9x) \bigg| \limits^9_8 - \int\limits^9_8 {\sqrt[3]{x - 9}} \, dx + \int\limits^{10}_9 {\sqrt[3]{x - 9}} \, dx - (x^2 - 9x) \bigg| \limits^9_8$ Evaluate [Integration Rule - FTC 1]: $\displaystyle A = \frac{-1}{2} - \int\limits^9_8 {\sqrt[3]{x - 9}} \, dx + \int\limits^{10}_9 {\sqrt[3]{x - 9}} \, dx - \frac{1}{2}$ Subtract: $\displaystyle A = -\int\limits^9_8 {\sqrt[3]{x - 9}} \, dx + \int\limits^{10}_9 {\sqrt[3]{x - 9}} \, dx - 1$ Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle A = -\int\limits^9_8 {(x - 9)^{\frac{1}{3}}} \, dx + \int\limits^{10}_9 {(x - 9)^{\frac{1}{3}}} \, dx - 1$

Step 4: Identify Variables

Identify variables for u-substitution.

u = x - 9

du = dx

Step 5: Find Area Pt. 2

[Integrals] U-Substitution: $\displaystyle A = -\int\limits^0_{-1} {u^{\frac{1}{3}}} \, du + \int\limits^1_0 {u^{\frac{1}{3}}} \, du - 1$ [Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = -(\frac{3}{4}u^{\frac{4}{3}}) \bigg| \limits^0_{-1} + (\frac{3}{4}u^{\frac{4}{3}}) \bigg| \limits^1_0 - 1$ Evaluate [Integration Rule - FTC 1]: $\displaystyle A = -(\frac{-3}{4}) + \frac{3}{4} - 1$ Simplify: $\displaystyle A = \frac{1}{2}$

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

Unit: Integrals - Area between 2 curves

Book: College Calculus 10e

Consider the following functions. (See attachment)
Find the area of the region.

Answer from: Quest

step-by-step explanation:

do you have a copy of the table?

Answer from: Quest

2,160 π $cm^{3}$

step-by-step explanation:

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Consider the following functions. (See attachment)
Find the area of the region.
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