Calculus: Analytic Geometry and Calculus, with Vectors

4.3 Properties of integrals 233

B6 -47,B7 = 0, B8 = -'96, and that IB2n1 is very large whenn is large. Some
books, particularly those that give a few formulas involving Bernoulli numbers
but do not treat Bernoulli functions, use notation which conflicts with the nota-
tion used above.
11 Prove that if f is integrable over the interval 0 < x 5 1, then

(1) nnonk if(n) =fo1f(x)dx.

Solution: To keep all of the bewitching mysticism of mysterious mathematics
out of our solution, let E be a given positive number. Choose a positive number
S such that

(2)

n

II

f(xk) Axk - fo 0

1

AX) dx I < e

k

whenever the sum is a Riemann sum formed for a partition P of the interval
0 < x < 1 for which JPJ < 3. Let N be an integer for which N > 2 and N >
1/S. Let n be an integer greater than N. Let P be a partition of the interval
0 < x < 1 into n equal subintervals each having length 1/n. Then xk = k/n
for each k. Let xk = xk so that xk = k/n for each k. Since Lxk = xk - xk_1 =
1/n for each k, we see that lPnI = 1/n. Since n > N, we have 1/n < 1/N and
hence 1/n < S. Therefore, IPnI < S and (2) holds when the sum is the Riemann
sum formed for the partition Pn. But for the partition P. we have xk = k/n
and Oxk = 1/n, so

(3) 1 f(xk)Axk= f1k1 =1 i f(kl.

k-1 k=1 n n n k=1 \n J)

It follows that

(^1) fCkl_ r1f(x) dx
nk=1 n 0
(4) < E
when n > N, and this gives the desired conclusion (1).
12 The basic formula (1) of Problem 11 has numerous quite astonishing
applications. Letting s be a nonnegative number and letting f(x) = x', use the
formula to prove that
(1) lim
18+28+38+ ... +n'1.
n-. w na+l 1 + s
Write the formulas to which this reduces when s = 0, T, 1, -, 2, and 3. Remark:
Textbooks that specialize in proofs by mathematical induction give the formulas
(2) 1 + 2 + 3 + ... +n =n(n+ 1)
2

(3) 12 + 22+ 32 + .. + n2 =n(n + 1)(2n+ 1)

6

(4) 13 +23+33 + ... +ns=n(n-I-1)2.

4