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CalculusCalculus (MindTap Course List)8th Edition

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Chapter 4.R, Problem 45E
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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

Solutions

Chapter
Section
Problem 1CC
Problem 2CC
Problem 3CC
Problem 4CC
Problem 5CC
Problem 6CC
Problem 7CC
Problem 8CC
Problem 9CC
Problem 1TFQ
Problem 2TFQ
Problem 3TFQ
Problem 4TFQ
Problem 5TFQ
Problem 6TFQ
Problem 7TFQ
Problem 8TFQ
Problem 9TFQ
Problem 10TFQ
Problem 11TFQ
Problem 12TFQ
Problem 13TFQ
Problem 14TFQ
Problem 15TFQ
Problem 16TFQ
Problem 17TFQ
Problem 18TFQ
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
Problem 45E
Problem 46E
Problem 47E
Problem 48E
Problem 49E
Problem 50E
Problem 51E
Problem 52E
Problem 53E
Problem 54E
Problem 55E
Problem 56E
Problem 57E
Problem 58E
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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

Use the Midpoint Rule with n = 6 to approximate ∫ 0 3 sin ( x 3 ) d x

To determine

To approximate:

The given integral ∫03sin⁡(x3)dx by using the midpoint rule with n=6

Explanation

1) Concept:

i) The midpoint rule: ∫abfxdx≈∑i=1nf(xi-)∆x=∆x[fx1-+·····+fxn-]

where ∆x=b-an  and  xi-=12xi-1+xi= midpoint of [xi-1,xi]

2) Given:

∫03sin⁡(x3)dx

3) Calculation:

Given that the integral is ∫03sin⁡(x3)dx. Applying the midpoint rule on it ∫03sin⁡(x3)dx≈∑i=16f(xi-)∆x Here

∆x=b-an=3-06=36=12

Now divide the interval [0,3] into 6 subintervals with endpoints 0 , 12, 1, 32, 2,52, 3

Therefore, the six subintervals are

0,12, 12, 1, 1,32, 32, 2 , 2,52, 52, 3

And the midpoints xi- of these intervals are

 Midpoint of 0,12=120+12=14

Midpoint of 12, 1=1212+1=34

Midpoint of 1,32=121+32=

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