   Chapter 5, Problem 57RE ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

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

To determine

To calculate: The approximate value of the integral function 03sin(x3)dx using Midpoint rule.

Explanation

Given information:

The integral function is 03sin(x3)dx.

Number of intervals n=6.

The interval varies from i=1,2,....,n.

Consider that the function as f(x)=sin(x3).

Show the midpoint Rule:

abf(x)dxi=1nf(x¯i)Δx=Δx(f(x¯1)+.......+f(x¯n))

Find the width of subinterval Δx using the formula.

Δx=ban (1)

Find the midpoint using the formula.

x¯i=12(xi1+xi) (2)

Here, width of the subintervals is Δx, the upper limit is b, the lower limit is a and the midpoint of (xi1,xi) is x¯i.

To find the width of the subintervals using Equation (1) as shown below.

Δx=306=0.5

Hence, the endpoints based on the width of subintervals are 0, 0.5, 1, 1.5, 2, 2.5, and 3.

To find the midpoint using Equation (2) as shown below.

Substitute 1 for i in Equation (2).

x¯1=12(x0+x1) (3)

Substitute 0 for x1 and 0.5 for x2 in Equation (3).

x¯1=12(0+0.5)=0.25

Similarly calculate the midpoints for the remaining intervals and tabulate the values

Shown the midpoint values as in Table (1).

 Interval, (i) Midpoint, x¯i 1 0.25 2 0.75 3 1.25 4 1.75 5 2.25 6 2.75

The expression to find the approximate value of the integral function using midpoint rule as shown below

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