Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 16.2 ITERATED INTEGRALS ET SECTION 15.2 ¤ 225
15. To calculate the estimates using a programmable calculator, we can use an algorithm
similartothatofExercise5.1.7[ET 5.1.7]. In Maple, we can define the function
f(x, y) = √ 1+xe −y (calling it f), load the student package, and then use the
command
middlesum(middlesum(f,x=0..1,m),
y=0..1,m);
to get the estimate with n = m 2 squares of equal size. Mathematica has no special
Riemann sum command, but we can define f andthenusenestedSum commands to
calculate the estimates.
n estimate
1 1.141606
4 1.143191
16 1.143535
64 1.143617
256 1.143637
1024 1.143642
17. If we divide R into mn subrectangles, R kdA≈ m
But f m
x ∗ ij,yij
∗ = k always and
i =1
points,
m
n
i =1 j =1
R kdA=
lim
i =1 j =1
n
f
x ∗ ij,yij ∗ ∆A for any choice of sample points x
∗
ij ,yij ∗ .
n
∆A = area of R =(b − a)(d − c). Thus, no matter how we choose the sample
j =1
f m
x ∗ ij,yij
∗ ∆A = k
i =1
m
n
m,n→∞ i =1 j =1
n
∆A = k(b − a)(d − c) and so
j =1
f x ∗ ij,yij ∗ ∆A = lim k m
m,n→∞ i =1
n
∆A =
j =1
lim
m,n→∞
k(b − a)(d − c) =k(b − a)(d − c).
16.2 Iterated Integrals ET 15.2
1.
3.
5.
7.
9.
x=5
5
0 12x2 y 3 dx =
12 x3
3 y3 =4x 3 y 3 x=5
x=0 =4(5)3 y 3 − 4(0) 3 y 3 = 500y 3 ,
x=0
1
0 12x2 y 3 dy =
12x 2 y 4 y=1
=3x 2 y 4 y=1
4
y=0 =3x2 (1) 4 − 3x 2 (0) 4 =3x 2
3
1
2
0
2
0
y=0
1 (1 + 4xy) dx dy = 3
0 1 x +2x 2 y x=1
dy = 3
(1 + 2y) dy = y + y 2 3
=(3+9)− (1 + 1) = 10
x=0 1 1
π/2
0
x sin ydydx= 2
0 xdx π/2
0
sin ydy [as in Example 5] =
2
1 1 (2x + (2x + y) 9
0 y)8 dx dy =
2 9
4
2
1
1
= 1 18
x
y + y x
dy dx =
0
2
0
x=1
x=0
[(2 + y) 9 − (0 + y) 9 ] dy = 1 18
x
2
2
2 π/2
− cos y =(2− 0)(0 + 1) = 2
0
0
dy [substitute u =2x + y ⇒ dx = 1 2 du]
(2 + y)
10
= 1
180 [(410 − 2 10 ) − (2 10 − 0 10 )] = 1,046,528
180
= 261,632
45
4
1
x ln |y| + 1 x · 1 y=2
2 y2 dx =
y=1
4
1
10
2
− y10
10
0
=8ln2+ 3 2 ln 4 − 1 2 ln 2 = 15 2 ln 2 + 3 ln 41/2 = 21
2 ln 2
x ln 2 + 3
dx = 1
2
2x
x2 ln 2 + 3 ln |x| 4
2 1