Mathematics, 29.05.2021 01:40 taijaababy

Consider the following. (See attachment) a) Find the area of the region
b) Use the integration capabilities of the graphing utility to verify you results. (Round your answer to one decimal place.)

Consider the following. (See attachment)

a) Find the area of the region
b) Use the integration ca

Answers: 1

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Answer from: almostnevercbbc

Area: 16

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Calculus

Derivatives

Derivative Notation

Integrals - Area under the curve

Trig Integration

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

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = 8sin(x) + sin(8x)$

$\displaystyle y = 0$

Bounds of Integration: 0 ≤ x ≤ π

Step 2: Find Area Pt. 1

Set up integral: $\displaystyle A = \int\limits^{\pi}_0 {[8sin(x) + sin(8x)]} \, dx$ Rewrite integral [Integration Property - Addition/Subtraction]: $\displaystyle A = \int\limits^{\pi}_0 {8sin(x)} \, dx + \int\limits^{\pi}_0 {sin(8x)} \, dx$ [1st Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 8\int\limits^{\pi}_0 {sin(x)} \, dx + \int\limits^{\pi}_0 {sin(8x)} \, dx$ [1st Integral] Integrate [Trig Integration]: $\displaystyle A = 8[-cos(x)] \bigg| \limits^{\pi}_0 + \int\limits^{\pi}_0 {sin(8x)} \, dx$ [1st Integral] Evaluate [Integration Rule - FTC 1]: $\displaystyle A = 8(2) + \int\limits^{\pi}_0 {sin(8x)} \, dx$ Multiply: $\displaystyle A = 16 + \int\limits^{\pi}_0 {sin(8x)} \, dx$

Step 3: Identify Variables

Identify variables for u-substitution.

u = 8x

du = 8dx

Step 4: Find Area Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 16 + \frac{1}{8}\int\limits^{\pi}_0 {8sin(8x)} \, dx$ [Integral] U-Substitution: $\displaystyle A = 16 + \frac{1}{8}\int\limits^{8\pi}_0 {sin(u)} \, du$ [Integral] Integrate [Trig Integration]: $\displaystyle A = 16 + \frac{1}{8}[-cos(u)] \bigg| \limits^{8\pi}_0$ [Integral] Evaluate [Integration Rule - FTC 1]: $\displaystyle A = 16 + \frac{1}{8}(0)$ Simplify: $\displaystyle A = 16$

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

Unit: Integration - Area under the curve

Book: College Calculus 10e

Answer from: Quest

If your asking how much altogether for 12 performance the answer will be 98,496.

Answer from: Quest

i need more details sorry

step-by-step explanation:

Asap will give brainlist. a. x = 11 b. x = 10 c. x = 14.4 d. x = 7

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Consider the following. (See attachment) a) Find the area of the region
b) Use the integratio...

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