Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 6.1 AREAS BETWEEN CURVES ¤ 267
35. From the graph, we see that the curves intersect at x =0and x = a ≈ 0.896,with
x sin(x 2 ) >x 4 on (0,a). So the area A of the region bounded by the curves is
A =
a
x sin(x 2 ) − x 4 dx = − 1 2 cos(x2 ) − 1 5 x5 a
0
0
= − 1 2 cos(a2 ) − 1 5 a5 + 1 2 ≈ 0.037
37. From the graph, we see that the curves intersect at
x = a ≈ −1.11,x= b ≈ 1.25, and x = c ≈ 2.86, with
x 3 − 3x +4> 3x 2 − 2x on (a, b) and 3x 2 − 2x >x 3 − 3x +4
on (b, c). So the area of the region bounded by the curves is
A =
=
b
a
b
a
(x 3 − 3x +4)− (3x 2 − 2x) c
dx + (3x 2 − 2x) − (x 3 − 3x +4) dx
(x 3 − 3x 2 − x +4)dx +
c
b
b
(−x 3 +3x 2 + x − 4) dx
= 1
4 x4 − x 3 − 1 2 x2 +4x b
a + − 1 4 x4 + x 3 + 1 2 x2 − 4x c
b ≈ 8.38
39. As the figure illustrates, the curves y = x and y = x 5 − 6x 3 +4x
enclose a four-part region symmetric about the origin (since
x 5 − 6x 3 +4x and x are odd functions of x). The curves intersect
at values of x where x 5 − 6x 3 +4x = x;thatis,where
x(x 4 − 6x 2 +3)=0. That happens at x =0and where
x 2 = 6 ± √ 36 − 12
=3± √ 6;thatis,atx = − 3+ √ 6, − 3 − √ 6, 0, 3 − √ 6,and 3+ √ 6.
2
The exact area is
√ 3+ √ 6
2 (x 5 − 6x 3 +4x) − x √ 3+ √ 6
dx =2 x 5 − 6x 3 +3x dx
0
0
√ 3− √ 6
√
=2 (x 5 − 6x 3 3+ √ 6
+3x) dx +2 (−x 5 +6x 3 − 3x) dx
0
CAS
= 12 √ 6 − 9
√
3− √ 6