# The Riemann sum with four subintervals for the function at the right endpoints.

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, Problem 2RE

(a)

To determine

Expert Solution

## Answer to Problem 2RE

The Riemann sum with four subintervals at the right endpoint is 1.25.

### Explanation of Solution

Given information:

The function as f(x)=x2x

The region lies between x=0 and x=2. Hence, the limits are a=0 and b=2.

Consider that the Number of rectangles as n=4.

The expression to find Riemann sum at the right endpoint (Rn) is shown below:

Rn=i=1nf(xi)Δx (1)

Here, the height of sample at the right endpoint is f(xi) and the width is Δx.

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 rectangles is n.

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

Δx=204=0.5

Calculate R4 using equation (1).

Substitute 4 for n in equation (1).

R4=i=14f(xi)Δx (3)

Substitute i=1,2,3,and 4 in equation (3).

R4=f(x1)Δx+f(x2)Δx+f(x3)Δx+f(x4)Δx (4)

At x1=0.5 the value of f(x1) is,

f(x1)=0.520.5=0.25

At x2=1 the value of f(x2) is,

f(x2)=121=0

At x3=1.5 the value of f(x3) is,

f(x3)=1.521.5=0.75

At x4=2 the value of f(x4) is,

f(x4)=222=2

Draw four rectangles at the right endpoints based on the functions as shown in Figure (1).

Substitute 0.5 for Δx, 0.25 for f(x1), 0 for f(x2), 0.75 for f(x3) and 2 for f(x4) in Equation (4).

R6=(0.25)×0.5+0×0.5+0.75×0.5+2×0.5=1.25

Therefore, Riemann sum with four subintervals at the right endpoint is 1.25.

(b)

To determine

Expert Solution

## Answer to Problem 2RE

The value of the integral is (23).

### Explanation of Solution

Given information:

The function as 02(x2x)dx.

The region lies between x=0 and x=2.

The expression to find the value of the integral using the definition of a definite integral is shown below:

02(x2x)dx=limni=1nf(xi)Δx (5)

Substitute 2 for b and 0 for a in Equation (2).

Δx=ban=20n=2n

To find the value of xi using the relation:

xi=a+iΔx (6)

Substitute 0 for a and 2n for Δx in equation (6).

xi=0+i2n=2in

Substitute (xi2xi) for f(xi) in equation (5).

02(x2x)dx=limni=1n(xi2xi)Δx (7)

Substitute 2n for Δx and 2in for xi in equation (7).

02(x2x)dx=limni=1n(4i2n22in)(2n)=limn(2n)(4n2i=1ni22ni=1ni)=limn(8n3×n(n+1)(2n+1)64n2×n(n+1)2)=limn(43×(n+1)n×(2n+1)n2×(n+1)n)

02(x2x)dx=limn(43×(1+1n)×(2+1n)2×(1+1n))=43×1×22×1=82×33=23

Therefore, the value of the integral using the definition of a definite integral is (23).

(c)

To determine

Expert Solution

## Answer to Problem 2RE

The value of the integral is (23).

### Explanation of Solution

Given information:

The function as 02(x2x)dx.

The region lies between x=0 and x=2.

The expression to find the value of the integral using the fundamental theorem is shown below:

02(x2x)dx=(13x312x2)02=(13×2312×22)(13×0312×02)=(832)0=23

Therefore, the value of the integral using the fundamental theorem is (23).

(d)

To determine

Expert Solution

## Answer to Problem 2RE

The geometric meaning of the integral is 02(x2x)dx=A1A2.

### Explanation of Solution

Given information:

The function as 02(x2x)dx.

The region lies between x=0 and x=2.

Draw the diagram based on the functions as in part (a) as shown in Figure (2).

Refer Figure (1).

Upper portion of the diagram is represented as A1 and lower portion of the diagram is represented as A2.

The expression to find the geometric value of the function from the diagram as shown below:

02(x2x)dx=A1A2

Therefore, the geometric mean of the integral (A1A2).

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