CALCULUS:EARLY TRANS.-SAPLINGPLUS
CALCULUS:EARLY TRANS.-SAPLINGPLUS
4th Edition
ISBN: 9781319279813
Author: Rogawski
Publisher: MAC HIGHER
Question
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Chapter 15, Problem 1CRE
To determine

Calculate the Riemann sum S2,3 for the given integral using two choices of sample points:

a) Lower-left vertex

b) Midpoint of rectangle

Then calculate the exact value of the double integral.

Expert Solution & Answer
Check Mark

Answer to Problem 1CRE

Solution:

(a) The Riemann sum S2,3 for the given double integral using lower-left vertices is 240.

(b)The Riemann sum S2,3 for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

Explanation of Solution

Given:

The integral: 1426x2ydxdy

Formulas:

Sm,n=i=1mj=1nf(xi,yj)ΔA

Where

ΔA=ΔxΔyΔx=bamandΔy=dcn

Calculations:

From the given integral, we can observe that 2x6and 1y4. Since our aim is to find S2,3, we need to divide the rectangle [2,6]×[1,3] into 2×3 subrectangles. The length and width of each subrectangle are calculated as follows:

Δx=622=42=2

Δy=413=33=1

Therefore, the area of each subrectangle is ΔA=ΔxΔy=21=2.

The subrectangles are shown in Image 1.

Image 1:

CALCULUS:EARLY TRANS.-SAPLINGPLUS, Chapter 15, Problem 1CRE , additional homework tip  1

(a) Using Lower-left vertex

Here, we use the lower-left vertices of each subrectangleto find the Riemann sum S2,3. Notice that the lower-left vertices are (2,1),(2,2),(2,3),(4,1),(4,2),and (4,3)and are shown in Image 2.

Image 2:

CALCULUS:EARLY TRANS.-SAPLINGPLUS, Chapter 15, Problem 1CRE , additional homework tip  2

Thus,

1426x2ydxdy=i=12j=13f(xi,yj)ΔA=[f(x1,y1)+f(x1,y2)+f(x1,y3)+f(x2,y1)+f(x2,y2)+f(x2,y3)]ΔA=[f(2,1)+f(2,2)+f(2,3)+f(4,1)+f(4,2)+f(4,3)]2=[221+222+223+421+422+423]2=[4+8+12+16+32+48]2=1202=240

(b) Using Midpoint of Rectangle:

Here, we use the midpoints of each subrectangle to find the Riemann sum S2,3. Notice that the midpoints are (3,1.5),(3,2.5),(3,3.5),(5,1.5),(5,2.5),and (5,3.5)and are shown in Image 3.

Image 3:

CALCULUS:EARLY TRANS.-SAPLINGPLUS, Chapter 15, Problem 1CRE , additional homework tip  3

Thus,

1426x2ydxdy=i=12j=13f(xi¯,yj¯)ΔA=[f(x1¯,y1¯)+f(x1¯,y2¯)+f(x1¯,y3¯)+f(x2¯,y1¯)+f(x2¯,y2¯)+f(x2¯,y3¯)]ΔA=[f(3,1.5)+f(3,2.5)+f(3,3.5)+f(5,1.5)+f(5,2.5)+f(5,3.5)]2=[321.5+322.5+323.5+521.5+522.5+523.5]2=[13.5+22.5+31.5+37.5+62.5+87.5]2=2552=510

To calculate the exact value of the integral:

1426x2ydxdy=14ydy26x2dx=y22]14x33]26=[422122][633233]=[812][7283]=1522083=520

Conclusion:

Thus,

(a) The Riemann sum S2,3 for the given double integral using lower-left vertices is 240.

(b)The Riemann sum S2,3 for the given double integral using midpoints is 510.

And the exact value of the double integral is 520.

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Chapter 15 Solutions

CALCULUS:EARLY TRANS.-SAPLINGPLUS

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