Single Variable Calculus: Concepts and Contexts, Enhanced Edition

4th Edition

ISBN: 9781337687805

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 5.9, Problem 8E

To determine

To find : the approximate value of integral with the specified value of n .

Expert Solution

Answer to Problem 8E

the approximate value of integral by trapezoidal rule is 0.083 .

Explanation of Solution

Given information :

The integral is ∫012sin(x2)dx,n=4 .

Formula used : trapezoidal rule:

∫abf(x)dx=Tn=Δx2[f(x0)+2f(x1)+2f(x2)+...+2f(xn−1)+f(xn)]

Where Δx=(b−a)n and xi=a+iΔx .

Calculation : by using trapezoidal rule,

∫012sin(x2)dx,T4

Since a=0,b=12 and n=4 .

So,

Δx=(b−a)nΔx=(12−0)4 [substitute]Δx=18 [simplify]

Since a=0,b=12 and n=4 , Δx=0.125

Now the value is

∫012sin(x2)dx=Δx2[f(x0)+2f(x1)+2f(x2)+2f(x3)+f(x4)]∫012sin(x2)dx=0.1252[f(0)+2f(0.125)+2f(0.25)+2f(0.375)+f(0.5)] [substitute]∫012sin(x2)dx=0.1252[sin(02)+2sin(0.1252)+2sin(0.252)+2sin(0.3752)+sin(0.52)] ∫012sin(x2)dx=0.1252[0+2·0.155+2·0.247+2·.139+.247]∫012sin(x2)dx=0.1252[1.329]∫012sin(x2)dx=0.083

So the approximate value of integral by trapezoidal rule is 0.083 .

To determine

To find : the approximate value of integral with the specified value of n .

Expert Solution

Answer to Problem 8E

the approximate value of integral by midpoint rule is 0.081 .

Explanation of Solution

Given information :

The integral is ∫012sin(x2)dx,n=4 .

Formula used:

Midpoint rule:

∫abf(x)dx=Mn=Δx[f(x¯1)+f(x¯2)+.....+f(x¯n)]Δx=b−anx¯i=12(xi−1+xi)=midpoint of [xi−1,xi]

Calculation: by using midpoint rule,

∫012sin(x2)dx,T4

Since a=0,b=12 and n=4 .

So,

Δx=(b−a)nΔx=(12−0)4 [substitute]Δx=18 [simplify]

The midpoint of the four subintervals are 116,316,516,716,916 .

So by the midpoint rule:

∫012sin(x2)dx=Δx2[f(x¯0)+f(x¯1)+f(x¯2)+f(x¯3)+f(x¯4)]∫012sin(x2)dx=0.125[f(116)+f(0.125)+f(0.25)+f(0.375)+f(0.5)] [substitute]∫012sin(x2)dx=0.125[sin(11622)+sin(0.1252)+sin(0.252)+sin(0.3752)+sin(0.52)] ∫012sin(x2)dx=0.125[0.1+0.31+0.494+..278+.247]∫012sin(x2)dx=0.125[1.329]∫012sin(x2)dx=0.081

So, the approximate value of integral by midpoint rule is 0.081 .

To determine

To find : the approximate value of integral with the specified value of n .

Expert Solution

Answer to Problem 8E

the approximate value by simpson’s rule is 0.085 .

Explanation of Solution

Given information :

The integral is ∫012sin(x2)dx,n=4 .

Formula used:

Simpson’s rule:

∫abf(x)dx=Sn=Δx3[f(x0)+4f(x1)+2f(x2)+...+4f(xn−1)+f(xn)]

Where n is even and Δx=(b−a)n .

Calculation: by using simpson’s rule,

∫012sin(x2)dx,T4

Since a=0,b=12 and n=4 .

So,

Δx=(b−a)nΔx=(12−0)4 [substitute]Δx=18 [simplify] By simpson’s rule,

∫012sin(x2)dx=Δx3[f(x0)+4f(x1)+2f(x2)+4f(x3)+f(x4)]∫012sin(x2)dx=0.1253[f(0)+4f(0.125)+2f(0.25)+4f(0.375)+f(0.5)] [substitute]∫012sin(x2)dx=0.1253[sin(02)+4sin(0.1252)+2sin(0.252)+4sin(0.3752)+sin(0.52)] ∫012sin(x2)dx=0.1253[0+4·0.155+2·0.247+4·.139+.247]∫012sin(x2)dx=0.1253[1.329]∫012sin(x2)dx=0.085

Thus, the approximate value by simpson’s rule is 0.085 .

Chapter 5.9, Problem 7E

Chapter 5.9, Problem 9E

Chapter 5 Solutions

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

Ch. 5.1 - Prob. 1E Ch. 5.1 - (a) Use six rectangles to find estimates of each...Ch. 5.1 - (a) Estimate the area under the graph of f(x) =...Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - The speed of a runner increased steadily during...Ch. 5.1 - Prob. 12E

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