Solução_Calculo_Stewart_6e

F.
356 ¤ CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION
TX.10
15. y =ln(1− x 2 ) ⇒ y 0 1
=
1 − x · (−2x) ⇒
2
2 dy
4x 2
1+ =1+
dx (1 − x 2 ) = 1 − 2x2 + x 4 +4x 2
= 1+2x2 + x 4
= (1 + x2 ) 2
⇒
2 (1 − x 2 ) 2 (1 − x 2 ) 2 (1 − x 2 ) 2
1+
So L =
2 dy 1+x
2 2
= = 1+x2
dx 1 − x 2 1 − x = −1+ 2
2 1 − x 2 [by division] = −1+ 1
1+x + 1
1 − x
1/2
0
[partial fractions].
−1+ 1
1+x + 1
dx = −x +ln|1+x| − ln |1 − x| 1/2
= − 1
1 − x
0 2 +ln3 − ln 1
2 2 − 0=ln3−
1
. 2
17. y = e x ⇒ y 0 = e x ⇒ 1+(y 0 ) 2 =1+e 2x .So
L =
1
0
e
1+e
2x
dx = du 1+u
2
u
√ 1+e 2
v
=
√
2 v 2 − 1 vdv
1
v = √ 1+u 2 ,so
v 2 =1+u 2 , vdv= udu
u = e x ,so
x =lnu, dx = du/u
√ 1+e 2
=
√
2
=
e
1
√
1+u
2
udu
u 2
1+ 1/2
v − 1 − 1/2
dv
v +1
= v + 1 2 ln v − 1 √ 1+e 2
= √ 1+e
v +1
2 + 1 √
√
2
2 ln 1+e2 − 1
√
1+e2 +1 − √ 2 − 1 √
2 − 1
2 ln √
2+1
= √ 1+e 2 − √ 2+ln √ 1+e 2 − 1 − 1 − ln √ 2 − 1
Or: Use Formula 23 for √ 1+u 2 /u du, or substitute u =tanθ.
19. y = 1 2 x2 ⇒ dy/dx = x ⇒ 1+(dy/dx) 2 =1+x 2 .So
L = 1
√
1+x2 dx =2 1
√
1+x2 dx [by symmetry]
−1
0
=2 √
1
2 2+
1
ln 1+ √ 2 − 0+ 1 ln 1 = √ 2+ln 1+ √ 2
2 2
21 =2 √
x 1+x2 + 1 ln x + √ 1+x 2
2 2 1
0
or substitute
x =tanθ
21. From the figure, the length of the curve is slightly larger than the hypotenuse
of the triangle formed by the points (1, 0), (3, 0),and(3,f(3)) ≈ (3, 15),
where y = f(x) = 2 3 (x2 − 1) 3/2 . This length is about √ 15 2 +2 2 ≈ 15,so
we might estimate the length to be 15.5.
y = 2 3 (x2 − 1) 3/2 ⇒ y 0 =(x 2 − 1) 1/2 (2x) ⇒ 1+(y 0 ) 2 =1+4x 2 (x 2 − 1) = 4x 4 − 4x 2 +1=(2x 2 − 1) 2 ,
so, using the fact that 2x 2 − 1 > 0 for 1 ≤ x ≤ 3,
L = 3
(2x2 − 1)
1
2 dx = 3
2x 2 − 1 dx = 3
1
1 (2x2 − 1) dx = 2
3 x3 − x 3
=(18− 3) − 2
− 1 = 46 =15.3.
1 3 3
23. y = xe −x ⇒ dy/dx = e −x − xe −x = e −x (1 − x) ⇒ 1+(dy/dx) 2 =1+e −2x (1 − x) 2 .Let
f(x) = 1+(dy/dx) 2 = 1+e −2x (1 − x) 2 .ThenL = 5
5 − 0
f(x) dx. Sincen =10, ∆x = = 1 .Now
0 10 2
L ≈ S 10 = 1/2
3 [f(0) + 4f 1
2
+2f(1) + 4f
3
2
+2f(2) + 4f
5
2
+2f(3) + 4f
7
2
+2f(4) + 4f
9
2
+ f(5)]
≈ 5.115840
The value of the integral produced by a calculator is 5.113568 (to six decimal places).