Page 111 - Schaum's Outline of Theory and Problems of Advanced Calculus

102 INTEGRALS [CHAP. 5

b
ð
2
x ½gðxÞ f ðxފ dx
l ¼
a
By idealizing the plane region, R,asa volume with uniform density one, the expression
2
2
½ f ðxÞ gðxފ dx stands in for mass and r has the coordinate representation x . (See Problem 5.25(b)
for more details.)

Solved Problems

DEFINITION OF A DEFINITE INTEGRAL
5.1. If f ðxÞ is continuous in ½a; bŠ prove that
n
b a X ð b
lim fa þ kðb aÞ ¼ f ðxÞ dx
n n
n!1 a
k¼1
Since f ðxÞ is continuous, the limit exists independent of the mode of subdivision (see Problem 5.31).
Choose the subdivision of ½a; bŠ into n equal parts of equal length x ¼ðb aÞ=n (see Fig. 5-1, Page 90). Let
k ¼ a þ kðb aÞ=n, k ¼ 1; 2; ... ; n. Then
n n ð b
X b a X kðb aÞ
lim f ð k Þ x k ¼ lim fa þ ¼ f ðxÞ dx
n n
k¼1 k¼1
n!1 n!1 a
n
k
1 X
5.2. Express lim f as a definite integral.
n!1 n n
k¼1
Let a ¼ 0, b ¼ 1inProblem 1. Then
n
k
1 X ð 1
lim f f ðxÞ dx
n!1 n n ¼
k¼1 0
ð 1
2
5.3. (a) Express x dx as a limit of a sum, and use the result to evaluate the given definite integral.
0
(b) Interpret the result geometrically.
2
2
2
2
(a)If f ðxÞ¼ x ,then f ðk=nÞ¼ ðk=nÞ ¼ k =n . Thus by Problem 5.2,
n 2 ð 1
1 X k 2
lim x dx
n!1 n n 2 ¼
k¼1 0
This can be written, using Problem 1.29 of Chapter 1,
1 1 1 2 n 1 þ 2 þ þ n
ð 2 2 2 ! 2 2 2
2
x dx ¼ lim þ þ þ ¼ lim
0 n!1 n n 2 n 2 n 2 n!1 n 3
¼ lim nðn þ 1Þð2n þ 1Þ
6n 3
n!1
1
¼ lim ð1 þ 1=nÞð2 þ 1=nÞ ¼
6 3
n!1
which is the required limit.
Note: By using the fundamental theorem of the calculus, we observe that
Ð 1 2 3 1 3 3
0 x dx ¼ðx =3Þj 0 ¼ 1 =3 0 =3 ¼ 1=3.
2
1
(b) The area bounded by the curve y ¼ x ,the x-axis and the line x ¼ 1isequal to .
3