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(b) Assuming unit density, the moment of inertia about the y2 axis of the typical rectangle shown above is k f k xk . Then Required moment of inertia lim 2
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f k xk lim 128 15
n X k 1
2 k xk
x2 4 x2 dx
5.26. Find the length of arc of the parabola y x2 from x 0 to x 1.
Required arc length 1 q 1 q 1 dy=dx 2 dx 1 2x 2 dx
1 p 1 2 p 1 4x2 dx 1 u2 du 2 0 0 p p p p 1 f1 u 1 u2 1 ln u 1 u2 gj2 1 5 1 ln 2 5 0 2 2 2 2 4
5.27. (a) (Disk Method) Find the volume generated by revolving the region of Problem 5.25 about the x-axis.
Required volume lim
n X k 1 n!1
y2 xk  k
4 x2 2 dx 512=15: p kx,
(b) (Disk Method) Find the volume of the frustrum of a paraboloid obtained by revolving f x 0 < a @ x @ b about the x-axis.
INTEGRALS b
[CHAP. 5
V  (c)
kx dx
k 2 b a2 : 2
(Shell Method) Find the volume obtained by orbiting the region of part (b) about the y-axis. Compare this volume with that obtained in part (b). b V 2 x kx dx 2kb3 =3
The solids generated by the two regions are di erent, as are the volumes.
MISCELLANEOUS PROBLEMS 5.28 If f x and g x are continuous in a; b , prove Schwarz s inequality for integrals:  b 2 b b f x g x dx @ f f x g2 dx fg x g2 dx
a a a
We have b
f f x g x g2 dx
f f x g2 dx 2
f x g x dx 2
fg x g2 dx A 0
for all real values of .
Hence by Problem 1.13 of 1, using (1) with b b b B2 f f x g2 dx; C f x g x dx A2 g x g2 dx;
a a a
we nd C @ A B , which gives the required result.
2 2 2
M 5.29. Prove that lim
M!1 0
dx  . x4 4 8
We have x4 4 x4 4x2 4 4x2 x2 2 2 2x 2 x2 2 2x x2 2 2x : According to the method of partial fractions, assume 1 Ax B Cx D x4 4 x2 2x 2 x2 2x 2 Then 1 A C x3 B 2A 2C D x2 2A 2B 2C 2D x 2B 2D so that A C 0, B 2A 2C D 0, 2A 2B 2C 2D 0, 2B 2D 1 Solving simultaneously, A 1, B 1, C 1, D 1. Thus 8 4 8 4 dx 1 x 2 1 x 2 dx dx 8 x2 2x 2 x4 4 8 x2 2x 2 1 x 1 1 dx 1 x 1 1 dx dx dx 2 2 2 8 x 1 1 8 x 1 1 8 x 1 1 8 x 1 2 1 1 1 1 1 ln x2 2x 2 tan 1 x 1 ln x2 2x 2 tan 1 x 1 C 16 8 16 8 Then M
M!1 0
! ( ) dx 1 M 2 2M 2 1 1  tan 1 M 1 tan 1 M 1 lim ln 8 8 8 x4 4 M!1 16 M 2 2M 2 1 dx , called an improper integral of the rst kind. Such integrals are considered x4 4 0 See also Problem 5.74.
We denote this limit by further in 12.
CHAP. 5]
INTEGRALS
x 5.30. Evaluate lim
0 x!0
sin t3 dt . x4
The conditions of L Hospital s rule are satis ed, so that the required limit is d x d sin t3 dt sin x3 sin x3 3x2 cos x3 1 dx 0 lim dx lim lim lim 3 d 4 x!0 x!0 4x x!0 d x!0 4 12x2 x 4x3 dx dx
5.31. Prove that if f x is continuous in a; b then
Let 
n X k 1
f x dx exists.
f k xk , using the notation of Page 91. Since f x is continuous we can nd numbers Mk
and mk representing the l.u.b. and g.l.b. of f x in the interval xk 1 ; xk , i.e., such that mk @ f x @ Mk . We then have m b a @ s
n X k 1
mk xk @  @
n X k 1
Mk xk S @ M b a
where m and M are the g.l.b. and l.u.b. of f x in a; b . The sums s and S are sometimes called the lower and upper sums, respectively. Now choose a second mode of subdivision of a; b and consider the corresponding lower and upper sums denoted by s 0 and S 0 respectively. We have must s0 @ S and S0 A s 2
To prove this we choose a third mode of subdivision obtained by using the division points of both the rst and second modes of subdivision and consider the corresponding lower and upper sums, denoted by t and T, respectively. By Problem 5.84, we have s @ t @ T @ S0 and s0 @ t @ T @ S 3
which proves (2). From (2) it is also clear that as the number of subdivisions is increased, the upper sums are monotonic decreasing and the lower sums are monotonic increasing. Since according to (1) these sums are also " bounded, it follows that they have limiting values which we shall call s and S respectively. By Problem " " 5.85, s @ S. In order to prove that the integral exists, we must show that s S. Since f x is continuous in the closed interval a; b , it is uniformly continuous. Then given any  > 0, we can take each xk so small that Mk mk < = b a . It follows that S s
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