Page 120 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 5] INTEGRALS 111
3
Ð x sin t dt
5.30. Evaluate lim 0 .
x!0 x 4
The conditions of L’Hospital’s rule are satisfied, so that the required limit is
d ð x 3 d
sin t dt 3 3 2 3
dx 0 sin x dx ðsin x Þ 3x cos x 1
lim ¼ lim ¼ lim ¼ lim ¼
x!0 d 4 x!0 4x 3 x!0 d 3 x!0 12x 2 4
dx ðx Þ dx ð4x Þ
ð b
5.31. Prove that if f ðxÞ is continuous in ½a; bŠ then f ðxÞ dx exists.
a
n
X
f ð k Þ x k ,using the notation of Page 91. Since f ðxÞ is continuous we can find numbers M k
Let ¼
k¼1
and m k representing the l.u.b. and g.l.b. of f ðxÞ in the interval ½x k 1 ; x k Š, i.e., such that m k @ f ðxÞ @ M k .
We then have
n n
X X
m k x k @ @
mðb aÞ @ s ¼ M k x k ¼ S @ Mðb aÞ ð1Þ
k¼1 k¼1
where m and M are the g.l.b. and l.u.b. of f ðxÞ in ½a; bŠ. The sums s and S are sometimes called the lower and
upper sums, respectively.
Now choose a second mode of subdivision of ½a; bŠ and consider the corresponding lower and upper
sums denoted by s and S respectively. We have must
0
0
s @ S and S A s ð2Þ
0
0
To prove this we choose a third mode of subdivision obtained by using the division points of both the first
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 @ S 0 and s @ t @ T @ S ð3Þ
0
which proves (2).
From (2)itisalso 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
s
bounded, it follows that they have limiting values which we shall call s and S respectively. By Problem
s
s
5.85, s @ S. Inorder to prove that the integral exists, we must show that s ¼ S.
Since f ðxÞ is continuous in the closed interval ½a; bŠ,itis uniformly continuous. Then given any > 0,
we can take each x k so small that M k m k < =ðb aÞ. It follows that
n n
X X
ðM k m k Þ x k < x k ¼
S s ¼ ð4Þ
b a
k¼1 k¼1
s
Now S s ¼ðS SÞþðS sÞþð s sÞ and it follows that each term in parentheses is positive and so is less
s
s
s
than by (4). In particular, since S s is a definite number it must be zero, i.e., S ¼ s. Thus, the limits of
the upper and lower sums are equal and the proof is complete.
Supplementary Problems
DEFINITION OF A DEFINITE INTEGRAL
ð 1
3
5.32. (a) Express x dx as a limit of a sum. (b)Use the result of (a)to evaluate the given definite integral.
0
(c)Interpret the result geometrically.
Ans. (b) 1
4
2 6
ð ð
2
5.33. Using the definition, evaluate (a) ð3x þ 1Þ dx; ðbÞ ðx 4xÞ dx.
Ans. (a)8, (b)9 0 3