12077-0130670227_ismsec1
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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.
C14S01.012: We integrate first with respect to x, then with respect to y:
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.
C14S01.014: We integrate first with respect to y, then with respect to x:
12
42
x2y3 dy dx =
12
1
4x2y4
42
dx =
12
60x2 dx =
20x3
12
= 180.
C14S01.015: We integrate first with respect to x, then with respect to y:
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.
C14S01.016: We integrate first with respect to x, then with respect to y:
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.
C14S01.017: We integrate first with respect to y, then with respect to x:
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.
C14S01.018: We integrate first with respect to y, then with respect to x:
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.
C14S01.019: We integrate first with respect to x, then with respect to y:
/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.
C14S01.021: We integrate first with respect to y, then with respect to x:
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.
C14S01.022: We integrate first with respect to x, then with respect to y:
10
22
x2ey dx dy =
10
1
3x3ey
22
dy
=
10
16
3ey dy =
16
3ey10
=16
3(e 1) 9.1641697517815746.
C14S01.023: We integrate first with respect to y, then with respect to x:
10
0
ex sin y dy dx =
10
ex cos y
0
dx
=
1
0
2ex dx =
2ex1
0
= 2e 2 3.436563656918.
C14S01.024: We integrate first with respect to x, then with respect to y:
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.
C14S01.027: We integrate first with respect to x, then with respect to y:
/20
e1
sin y
xdx dy =
/20
(ln x)sin y
e1
dy
=
/20
sin y dy =
cos y
/20
= 0 (1) = 1.
C14S01.028: We integrate first with respect to y, then with respect to x:
e1e1
1
xydy dx
=e1 ln y
xe1
dx=e1
1
xdx
=
lnxe
1 = 1 0 = 1.
C14S01.029: We integrate first with respect to x, then with respect to y:
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.
C14S01.030: We integrate first with respect to y, then with respect to x:
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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C14S01.032: The first evaluation yields
/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.
C14S01.033: The first evaluation yields
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 .
C14S01.034: The first evaluation yields
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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