BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 5.2, Problem 12E
To determine

To evaluate: The integral function 0πxsin2xdx using midpoint Rule.

Expert Solution

Answer to Problem 12E

The approximate value of the integral function 0πxsin2xdx is2.4674 by using midpoint rule.

Explanation of Solution

Given information:

The integral function is 0πxsin2xdx and n=4.

The Expression for midpointrule is shown below:

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

Find the width (Δx) using the relation:

Δx=ban (2)

Here, the upper limit is b, the lower limit is a, and the number of intervals is n.

Substitute π for b, 0 for a, and 4 for n in Equation (2).

Δx=π04=π4

Calculate the value of right end points (xi) using the relation:

xi=a+iΔx (3)

Calculate the value of left end points (xi1) by rearranging Equation (3).

Substitute (i1) for i in Equation (3)

xi1=a+(i1)Δx (4)

Calculate the value of mid points (x¯i) using the relation:

x¯i=12(xi1+xi) (5)

Substitute a+iΔx for xi and a+(i1)Δx for xi1 in Equation (5).

x¯i=12[a+(i1)Δx+a+iΔx]=12(a+iΔxΔx+a+iΔx)=12×[2(a+iΔx)Δx]=(a+iΔx)Δx2 (6)

Calculate the value of x¯1 using Equation (6).

Substitute 0 for a, 1 for i,and π4 for Δx in Equation (6).

x¯1=(a+iΔx)Δx2=(0+1×π4)12×π4=π4π8=π8

Calculate the value of x¯2 using Equation (6).

Substitute 0 for a, 2 for i,and π4 for Δx in Equation (6).

x¯2=(0+2×π4)12×π4=2π4π8=3π8

Calculate the value of x¯3 using Equation (6).

Substitute 0 for a, 3 for i,and π4 for Δx in Equation (6).

x¯3=(a+iΔx)Δx2=0+(3×π4)(12×π4)=3π4π8=5π8

Calculate the value of x¯4 using Equation (6).

Substitute 0 for a, 3 for i,and π4 for Δx in Equation (6).

x¯4=(a+iΔx)Δx2=(0+4×π4)12×π4=ππ8=7π8

Consider f(x)=xsin2x.

Rearrange the expression f(x)=xsin2x in terms of f(x¯i) as shown below.

Substitute x¯i for x.

f(x¯i)=x¯isin2x¯i (7)

Calculate the value of f(x¯1) using Equation (7).

Substitute π8 for x¯1 in Equation (7).

f(x¯1)=x¯1sin2x¯1=π8×sin2(π8)

Calculate the value of f(x¯2) using Equation (7).

Substitute 3π8 for x¯2 in Equation (7).

f(x¯2)=x¯2sin2x¯2=3π8×sin2(3π8)

Calculate the value of f(x¯3) using Equation (7).

Substitute 5π8 for x¯3 in Equation (7).

f(x¯3)=x¯3sin2x¯3=5π8×sin2(5π8)

Calculate the value of f(x¯4) using Equation (7).

Substitute 7π8 for x¯4 in Equation (7).

f(x¯4)=x¯4sin2x¯4=7π8×sin2(7π8)

Substitute xsin2x for f(x),π4 for Δx,π8×sin2(π8) for f(x¯1), 3π8×sin2(3π8) for f(x¯2), 5π8×sin2(5π8) for f(x¯3),7π8×sin2(7π8) for f(x¯4), 0 for a, and π for b in Equation (1).

0πxsin2xdx=Δx[f(x¯1)+f(x¯2)+f(x¯3)+f(x¯4)]=π4×[π8×sin2(π8)+3π8×sin2(3π8)+5π8×sin2(5π8)+7π8×sin2(7π8)]=π4×[π8×sin2(π8)+3π8×sin2(3π8)+5π8×sin2(5π8)+7π8×sin2(7π8)]=π4×[π8×(0.1464)+3π8×(0.8536)+5π8×(0.8536)+7π8×(0.1464)]

=π232×[(0.1464)+3(0.8536)+5(0.8536)+7(0.1464)]=π232×[0.1464+2.5607+4.2678+1.0251]=0.3084×8=2.4674

Thus, the value of integral function 0πxsin2xdx is 2.4674.

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!