# The integral function ∫ 0 π x sin 2 x d x using midpoint Rule.

### Single Variable Calculus: Concepts...

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

### 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

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.

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