# 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 11E
To determine

Expert Solution

## Answer to Problem 11E

The integral value is0.9071.

### Explanation of Solution

Given information:

The integral function is 02xx+1dx,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 2 for b, 0 for a and 5 for n in Equation (1).

Δx=205=25

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 the 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 25 for Δx in Equation (4).

x¯i=(a+iΔx)Δx2=(0+1×25)12×(25)=2515=15

Calculate x¯2 using Equation (4).

Substitute 0 for a, 2 for i,and 25 for Δx in Equation (4).

x¯i=(a+iΔx)Δx2=(0+2×25)12×(25)=4515=35

Calculate x¯3 using Equation (4).

Substitute 0 for a, 3 for i,and 25 for Δx in Equation (4).

x¯i=(a+iΔx)Δx2=(0+3×25)12×(25)=6515=1

Calculate x¯4 using Equation (4).

Substitute 0 for a, 4 for i,and 25 for Δx in Equation (4)

x¯i=(a+iΔx)Δx2=(0+4×25)12×(25)=8515=75

Calculate x¯5 using Equation (4).

Substitute 0 for a, 5 for i,and 25 for Δx in Equation (4)

x¯i=(a+iΔx)Δx2=(0+5×25)12×(25)=215=95

Compare the integral function 02xx+1dx with Equation (1).

f(x)=xx+1

Calculate f(x¯i) using the Equation:

Substitute x¯i for x.

f(x¯i)=x¯ix¯i+1 (5)

Calculate f(x¯1) using Equation (5).

Substitute 15 for x¯1 in the Equation (5).

f(x¯i)=x¯ix¯i+1=(1515+1)=15×56=16

Calculate f(x¯2) using Equation (5).

Substitute 35 for x¯2 in the Equation (5)

f(x¯i)=x¯ix¯i+1=(3535+1)=35×58=38

Calculate f(x¯3) using Equation (5).

Substitute 1 for x¯3 in the Equation (5)

f(x¯i)=x¯ix¯i+1=(11+1)=12

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

Substitute 75 for x¯4 in the Equation (5)

f(x¯i)=x¯ix¯i+1=(7575+1)=75×512=712

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

Substitute 95 for x¯4 in the Equation (5)

f(x¯i)=x¯ix¯i+1=(9595+1)=95×514=914

Calculate 02xx+1dx using the Equation (1):

Substitute xx+1 for f(x),x¯ix¯i+1 for f(x¯i), 25 for Δx,16 for f(x¯1), 38 for f(x¯2), 12 for f(x¯3),and 712 for f(x¯4), 914 for f(x¯5), 0 for a and 2 for b in Equation (1).

02xx+1dx=Δx[f(x¯1)+f(x¯2)+f(x¯3)+f(x¯4)+f(x¯5)]=25[16+38+12+712+914]=25×[0.1667+0.3750+0.5+0.5833+0.6429]=25×(2.2679)

02xx+1dx=0.9071

Thus, the value of 02xx+1dx is 0.9071.

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