πExpress the surface area of the surface obtained by rotating the graph of y = 4 sin (x) for 0 ≤ x ≤not evaluate).Choose the correct integral.π/6O 2П4 sin x- 1 + 16 cos² (x) dxO2x ["4 sin x (1 + 4 cos (x)) dxπ/6Osin x√/1 + 16 cos² (x) dx10.20π/6O 2π6.1²√₁+ 16 cos² (x) dxπ/6O1 + 16 cos² (x) dxπ/656as an integral (but do

Recall: Surface area SA=∫ab2πy1+dydx2dx We have, y=4sin(x), a=0, b=π6

Answered: T Express the surface area of the…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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a. Explain why 0 ≤ x²arctan(x) ≤ (pi*x²)/4 for all 0 ≤ x ≤ 1. b. Use the properties of the integrals to show that the value of the integral lower bound is 0, higher bound is 1 and the integral is x² arctan(x) dx lies on the interval [0,pi/12]
a) Use the Trapezoid Rule to estimate the following integral. Use n=5 integral of (e^(x^2)) dx from 0 to 1 b) Use the Error Bound Formula for the Trapezoid Rule to find a best estimate for the error involved in part a
a) Use the reduction formula twice to find the value of this integral: integral tan^(5)xdx Reduction formula:integral tan^(m)xdx=(tan^(m-1)x)/m-1 - integral tan^(m-2)xdx. b) Use the Product to Sum Identities and integration to find the Area of the Q1 region trapped between the graph of the function f(x) and the x – axis over the given interval. Sketch the region first and estimate its’ size before integrating in each case. 1) f(x)=4sin(3x)cosx, I: [0, pi/3] 2) f(x)= 6cos(2x)cosx, I: [0, pi/4]
1.Using n=8 approximate the value of Integral with limits 4 and 0 cos (1+√x dX using Simpson's rule. 2.Using n=8 approximate. the value of integral with limits 2 and - 1 √e^-x2 +1 dx, using Trapezoid rule
(a) Use the Trapezoidal Rule, with n=5 to approximate the integral ∫109cos(4x)dx T5= (b) The actual value of ∫109cos(4x)dx= (c) The error involved in the approximation of part (a) is ET=∫109cos(4x)dx−T5= (d) The second derivative f″(x)=The value of KK = max |f″(x)| on the interval [0, 1] = . (e) Find a sharp upper bound for the error in the approximation of part (a) using the Error Bound Formula |ET| ≤K(b−a^31/2n^2 (f) Find the smallest number of partitions nn so that the approximation TnTn to the integral is guaranteed to be accurate to within 0.001. n=

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T Express the surface area of the surface obtained by rotating the graph of y = 4 sin (x) for 0 ≤ x ≤ 7 as an integral (but do not evaluate).