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Section 14.1

C14S01.001: Part (a):

f(1, 2) 1 + f(2, 2) 1 + f(1, 1) 1 + f(2, 1) 1 + f(1, 0) 1 + f(2, 0) 1 = 198.

Part (b):

f(2, 1) 1 + f(3, 1) 1 + f(2, 0) 1 + f(3, 0) 1 + f(2, 1) 1 + f(3, 1) 1 = 480.

The average of the two answers is 339, fairly close to the exact value 312 of the integral.

C14S01.002: Part (a):

f(1, 1) 1 + f(2, 1) 1 + f(1, 0) 1 + f(2, 0) 1 + f(1, 1) 1 + f(2, 1) 1 = 144.

Part (b):

f(2, 2) 1 + f(3, 2) 1 + f(2, 1) 1 + f(3, 1) 1 + f(2, 0) 1 + f(3, 0) 1 = 570.The average of the two answers is 357, fairly close to the exactly value 312 of the integral. The computations

shown here can be automated in computer algebra systems. For example, in Mathematica3.0, after defining

f(x, y) = 4x3 + 6xy2, you could proceed as follows.

x[i ] := i + 1; y[j ] := j - 2; deltax = x[1] - x[0]; deltay = y[1] - y[0];

( Part (a): ) xstar[i ] := x[i-1]; ystar[j ] := y[j]Sum[ Sum[ f[xstar[i],ystar[j]]deltaxdeltay, {j, 1, 3}, {i, 1, 2} ]

144

(

Part (b):

) xstar[i ] := x[i]; ystar[j ] := y[j-1]

Sum[ Sum[ f[xstar[i],ystar[j]]deltaxdeltay, {j, 1, 3}, {i, 1, 2} ]570

The idea is that to work another such problem, all you need to do is redefine f, xstar and ystar, and the

limits on i and j.

C14S01.003: We omit x and y from the computation because each is equal to 1.

f1

2, 12

+ f

3

2, 12

+ f

1

2, 32

+ f

3

2, 32

= 8.

This is also the exact value of the iterated integral.

C14S01.004: We omit x and y from the computation because each is equal to 1.

f1

2, 12

+ f

3

2, 12

+ f

1

2, 32

+ f

3

2, 32

= 4.

This is also the exact value of the iterated integral. In a Mathematica 3.0 solution similar to the one in

Problem 2, we would use

xstar[i ] := (x[i] + x[i-1])/2; ystar[j ] := (y[j] + y[j-1])/2

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C14S01.005: The Riemann sum is

f(2, 1) 2 + f(4, 1) 2 + f(2, 0) 2 + f(4, 0) 2 = 88.

The true value of the integral is416

3 138.666666666667.

C14S01.006: We omit x = 1 and y = 1 from the computation.

f(1, 1) + f(2, 1) + f(1, 2) + f(2, 2) + f(1, 3) + f(2, 3) = 43.

The true value of the integral is 26. The midpoint approximation gives the very close Riemann sum 25.

C14S01.007: We factor out of each term in the sum the product x y = 14

2. The Riemann sum then

takes the form

1

42 f 1

4, 1

4

+ f3

4, 1

4

+ f1

4, 3

4

+ f3

4, 3

4

= 12

2 4.935.

The true value of the integral is 4.

C14S01.008: We factor out of each term in the sum the product x y = 16

. The Riemann sum then

takes the form

1

6 f1

4, 16

+ f3

4, 16

+ f1

4, 12

+ f3

4, 12

+ f1

4, 56

+ f3

4, 56

= 12

1.571.

Mathematica3.0 reports that the true value of the integral is

- 14 CosIntegral[4] +14 (EulerGamma + Log[4])

and when we asked for a numerical value with the command N[%], it returned the approximation

0.77858913775068568

C14S01.009: Because f(x, y) = x2y2 is increasing in both the positive x-direction and the positive

y-direction on [1, 3] [2, 5], L M U.

C14S01.010: Because f(x, y) =

100 x2 y2 is decreasing in both the positive x-direction and thepositive y-direction on [1, 4] [2, 5], U M L.

C14S01.011: We integrate first with respect to x, then with respect to y:

2

0

4

0

(3x + 4y) dx dy = 2

0 3

2x2 + 4xy

4

0

dy = 2

0

(24 + 16y) dy = 24y + 8y22

0

= 80.


30

20

x2y dx dy =

30

1

3x3y

20

dy =

30

8

3y dy =

4

3y230

= 12.

C14S01.013: We integrate first with respect to y, then with respect to x:

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21

31

(2x 7y) dy dx =21

2xy 7

2y231

dx =

21

(4x 28) dx =

2x2 28x21

= 48 30 = 78.


12

42

x2y3 dy dx =

12

1

4x2y4

42

dx =

12

60x2 dx =

20x3

12

= 180.


30

30

(xy + 7x + y) dx dy =

30

xy +

7

2x2 +

1

2x2y

30

dy

=

30

3

2(5y + 21)

dy =

1

4(15y2 + 126y)

30

=513

4= 128.25.


20

42

(x2y2 17) dx dy =20

1

3x3y2 17x

42

dy

=

20

2

3(28y2 51) dy =

2

9(28y3 153y)

20

= 1649

18.222222222222.


2

1

2

1

(2xy2

3x2y) dy dx =

2

1 2

3xy3

3

2x2y2

2

1

dx

=

21

3

2(4x 3x2) dx =

3x2 3

2x321

= 0 92

= 92

= 4.5.


31

1

3

(x3y xy3) dy dx =31

1

2x3y2 1

4xy4

1

3

dx

= 3

1

(20x 4x3) dx = 10x2 x4

3

1

= 9 9 = 0.


/20

/20

sin x cos y dx dy =

/20

cos x cos y

/20

dy

=

/20

cos y dy =

sin y

/20

= 1 0 = 1.

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C14S01.020: This is merely Problem 19 with x and y interchanged, so the answer should be the same.

/20

/20

cos x sin y dy dx =

/20

cos x cos y

/20

dx

=/20

cos x dx =

sin x/20

= 1 0 = 1.


10

10

xey dy dx =

10

xey

10

dx

=

10

(ex x) dx =

1

2(e 1)x2

10

=1

2(e 1) 0.8591409142295226.


10

22

x2ey dx dy =

10

1

3x3ey

22

dy

=

10

16

3ey dy =

16

3ey10

=16

3(e 1) 9.1641697517815746.


10

0

ex sin y dy dx =

10

ex cos y

0

dx

=

1

0

2ex dx =

2ex1

0

= 2e 2 3.436563656918.


10

10

ex+y dx dy =

10

ex+y

10

dy =

10

ey+1 ey dy

=

ey+1 ey

10

= (e2 e) (e 1) = (e 1)2 2.9524924420125598.

C14S01.025:We integrate first with respect to x, then with respect to y:0

0

(xy + sin x) dx dy =

0

1

2x2y cos x

0

dy =

0

2 +

1

22y

dy

=

2y +

1

42y2

0

=1

4

4 + 8

30.635458065680.C14S01.026: We integrate first with respect to x, then with respect to y:

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/20

/20

(y 1) cos x dx dy =/20

(y 1)sin x

/20

dy =

/20

(y 1) dy

= 1

2y2 y

/2

0

=1

8(2 4) 0.3370957766587268.


/20

e1

sin y

xdx dy =

/20

(ln x)sin y

e1

dy

=

/20

sin y dy =

cos y

/20

= 0 (1) = 1.


e1e1

1

xydy dx

=e1 ln y

xe1

dx=e1

1

xdx

=

lnxe

1 = 1 0 = 1.


10

10

1

x + 1+

1

y + 1

dx dy =

10

x

y + 1+ ln(x + 1)

10

dy =

10

1

y + 1+ ln 2

dy

=

ln(y + 1) + y ln 2

10

= 2l n2 0 = 2ln2 1.3862943611198906.


2

1

3

1

x

y+

y

x

dy dx =

2

1

y2

2x+ x ln y

3

1

dx =

2

1

4

x+ x ln 3

dx

=

1

2x2 ln3 + 4ln x

21

= 4ln 2 +3

2ln 3 4.4205071552419458.

C14S01.031: The first evaluation yields

22

11

(2xy 3y2) dx dy =22

x2y 3xy2

11

dy

=22(6y

2

) dy = 2y3

2

2 = 16 16 = 32.The second yields

11

22

(2xy 3y2) dy dx =11

xy2 y3

22

dx

=

11

(16) dx =

16x11

= 16 16 = 32.

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/2/2

0

sin x cos y dx dy =

/2/2

cos x cos y

0

dy

=/2/2 2cos y dy =

2sin y

/2/2 = 2 (2) = 4.

The second yields

0

/2/2

sin x cos y dy dx =

0

sin x sin y

/2/2

dx

=

0

2sin x dx =

2cos x

0

= 2 (2) = 4.


2

1

1

0

(x + y)1/2 dx dy =

2

1

2

3(x + y)3/2

1

0

dy =

2

1

2

3(y + 1)3/2 2

3y3/2

dy

=

4

15(y + 1)5/2 4

15y5/2

21

=4

15

9

3 8

2 + 1

1.406599671769.

The second yields

10

21

(x + y)1/2 dy dx =

10

2

3(x + y)3/2

21

dx =

10

2

3(x + 2)3/2 2

3(x + 1)3/2

dx

= 415

(x + 2)5/2

4

15(x + 1)5/2

1

0

=4

1593 8

2 + 1 .


ln 30

ln 20

ex+y dx dy =

ln 30

ex+y

ln 20

dy =

ln 30

(2ey ey) dy =

eyln 30

= 3 1 = 2.

The second yields

ln 20

ln 30

ex+y dy dx =

ln 20

ex+y

ln 30

dx =

ln 20

(3ex ex) dx =

2exln 20

= 4 2 = 2.

C14S01.035: We may assume that n 1 and, if you wish, even that n is a positive integer. Then

10

10

xnyn dx dy =

10

xn+1yn

n + 1

10

dy =

10

yn

n + 1dy =

yn+1

(n + 1)2

10

=1

(n + 1)2.

Therefore

limn

10

10

xnyn dx dy = limn

1

(n + 1)2= 0.

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C14S01.036: Note that whatever the choice of (xi , yi ), f(x

i , y

i ) = k, and hence f(x

i , y

i ) Ai is equal

to the product of k and the area a(Ri) of Ri for each i. Hence

ni=1

f(xi , yi ) Ai =

ni=1

k a(Ri) = k

ni=1

a(Ri)

= k a(R) = k(b a)(d c).

C14S01.037: Let a(R) denote the area ofR. If 0 x and 0 y , then 0 f(x, y) sin 12

= 1.

Hence every Riemann sum lies between 0 a(R) and 1 a(R). Therefore

0

0

0

sin

xy dx dy a(R) = 2 9.869604401.

The exact value of the integral is

0

0

sin

xy dx dy = 4

0

sin t

tdt 7.4077482079298646814442134806319654533832.

C14S01.038:The corresponding relation between Riemann sums is

ni=1

cf(xi , yi ) Ai = c

ni=1

f(xi , yi ) Ai.

C14S01.039: The corresponding relation among Riemann sums is

ni=1

[f(xi , yi ) + g(x

i , y

i )] Ai =

ni=1

f(xi , yi ) Ai

+

ni=1

g(xi , yi ) Ai

.

C14S01.040: The corresponding relation between Riemann sums is this: If f(x, y) g(x, y) at each

point of R, thenni=1

f(xi , yi ) Ai

ni=1

g(xi , yi ) Ai.

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