# The integral 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 10E
To determine

## To evaluate: The integral using Midpoint Rule.

Expert Solution

The value of 01(x3+1)dx is 1.1097.

### Explanation of Solution

Given information:

The integral function 01(x3+1)dx,n=5.

Apply Midpoint Rule.

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

Find the width (Δx) using the relation:

Δx=ban

Substitute 1 for b, 0 for a, and 5 for n in Equation (1).

Δx=105=15=0.2

Calculate right end points xi using the relation:

xi=a+iΔx (2)

Calculate left end points xi1 using Equation (2)

Substitute (i1) for i in Equation (2)

xi1=a+(i1)Δx (3)

Calculate mid points using the relation:

x¯i=12(xi1+xi)

Substitute a+iΔx for xi and [a+(i1)Δx] for xi1

x¯i=12(xi1+xi)=12[a+(i1)Δx+a+iΔx]=12(a+iΔxΔx+a+iΔx)=12×[2(a+iΔx)Δx]

x¯i=(a+iΔx)Δx2 (4)

Calculate x¯1 using Equation (4)

Substitute 0 for a, 1 for i,and 0.2 for Δx in Equation (4).

x¯i=(a+iΔx)Δx2=(0+1×0.2)12×(0.2)=0.20.1=0.1

Calculate x¯2 using Equation (4).

Substitute 0 for a, 2 for i,and 0.2 for Δx in Equation (4).

x¯i=(a+iΔx)Δx2=(0+2×0.2)0.22=0.40.1=0.3

Calculate x¯3 using Equation (4)

Substitute 0 for a, 3 for i,and 0.2 for Δx in Equation (4)

x¯i=(a+iΔx)Δx2=(0+3×0.2)0.22=0.60.1=0.5

Calculate x¯4 using Equation (4)

Substitute 0 for a, 4 for i,and 0.2 for Δx in Equation (4)

x¯i=(a+iΔx)Δx2=(0+4×0.2)0.22=0.80.1=0.7

Calculate x¯5 using Equation (4)

Substitute 0 for a, 5 for i,and 0.2 for Δx in Equation (4)

x¯i=(a+iΔx)Δx2=(0+5×0.2)0.22=10.1=0.9

Compare the integral function 01(x3+1)dx with Equation (1).

f(x)=(x3+1)

Calculate f(xi) using the Equation:

Substitute x¯i for x.

f(x¯i)=(x¯i3+1) (5)

Calculate f(x¯1) using Equation (5).

Substitute 0.1 for x¯1 in the Equation (5)

f(x¯i)=(x¯i3+1)=((0.1)3+1)=1.001=1.0005

Calculate f(x¯2) using Equation (5).

Substitute 0.3 for x¯2 in the Equation (5)

f(x¯i)=(x¯i3+1)=((0.3)3+1)=1.027=1.0134

Calculate f(x¯3) using Equation (5).

Substitute 0.5 for x¯3 in the Equation (5)

f(x¯i)=(x¯i3+1)=((0.5)3+1)=1.125=1.0607

Calculate f(x¯4) using the Equation (5).

Substitute 0.7 for x¯4 in the Equation (5)

f(x¯i)=(x¯i3+1)=((0.7)3+1)=1.343=1.1589

Calculate f(x¯5) using the Equation (5).

Substitute 0.9 for x¯4 in the Equation (5)

f(x¯i)=(x¯i3+1)=((0.9)3+1)=1.729=1.3149

Calculate 01(x3+1)dx using the Equation (1):

Substitute (x3+1) for f(x),(x¯i3+1) for f(x¯i), 0.2 for Δx,1.0005 for f(x¯1), 1.0134 for f(x¯2), 1.0607 for f(x¯3),and 1.1589 for f(x¯4), 1.3149 for f(x¯5), 0 for a and 1 for b in Equation (1).

01(x3+1)dx=Δx[f(x¯1)+f(x¯2)+f(x¯3)+f(x¯4)+f(x¯5)]=0.2[1.0005+1.0134+1.0607+1.1589+1.3149]=0.2×5.5484=1.1097

Thus, the value of 01(x3+1)dx is 1.1097.

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