1. Some Examples Definition 1.1. Let a be any positive real number. We define
a0= 1.
Example 1.1. Let b > 1 and f : [1, b] → R be the function f (x) = 1/x for 1 ≤ x ≤ b. In this example, we are going to show that f is Riemann integrable (under certain assumption).
Let n be any natural number and h = √n
b. Set xi= hi for 0 ≤ i ≤ n. Then Pn= {xi: 0 ≤ i ≤ n}
is a partition of [1, b]. We let Miand mibe the supremum and infimum of f on [xi−1, xi] respectively.
For each 1 ≤ i ≤ n, we have
Mi = 1 xi−1
= 1
hi−1, mi= 1 xi
= 1 hi.
The upper Riemann sum U (f, Pn) and the lower Riemann sum of L(f, Pn) are given by U (f, Pn) =
n
X
i=1
Mi∆xi=
n
X
i=1
1
hi−1hi−1(h − 1) = n(h − 1) U (f, Ln) =
n
X
i=1
mi∆xi=
n
X
i=1
1
hihi−1(h − 1) =n(h − 1)
h .
By definition,
n(h − 1)
h ≤
Z b 1
1 xdx ≤
Z b 1
1
xdx ≤ n(h − 1).
More precisely,
(1.1) n(√n
b − 1)
√n
b ≤
Z b 1
1 xdx ≤
Z b 1
1
xdx ≤ n(√n b − 1).
Suppose we know the following facts:
(1) lim
n→∞n(n
√
b − 1) exists. We set I = lim
n→∞n(n
√ b − 1).
(2) If lim
n→∞an= a and lim
n→∞bn= b with b 6= 0, then
n→∞lim an
bn = a b. (3) lim
n→∞
√n
b = 1.
Then
n→∞lim n(√n
b − 1)
√n
b = I.
Taking n → ∞ of (1.1), we obtain that I ≤
Z b 1
1 xdx ≤
Z b 1
1
xdx ≤ I.
Hence
Z b 1
1 xdx =
Z b 1
1
xdx = I.
Therefore f (x) = 1/x, 1 ≤ x ≤ b, is Riemann integrable and Z b
1
1
xdx = I.
Remark. We will see (define) that
ln b = Z b
1
1 xdx.
1
2
Example 1.2. Let m ∈ N be a natural number and 0 < a < b. Let f : [a, b] → R be the function f (x) = xm. Show that f is Riemann integrable and
Z b a
xmdx = bm+1− am+1 m + 1 .
Let h = pb/a and xn i = ahi for 0 ≤ i ≤ n. Then Pn= {xi: 0 ≤ i ≤ n} forms a partition of [a, b].
Let Mi and mi be the supremum and infimum of f on [xi−1, xi] respectively. Then for 1 ≤ i ≤ m, Mi= xmi , mi= xmi−1.
Hence the upper Riemann sum U (f, Pn) and the lower Riemann sum L(f, P ) are given respectively by
U (f, Pn) =
n
X
i=1
(ahi)mahi−1(h − 1) =am+1(h − 1) h
n
X
i=1
h(m+1)i= am+1hm+1 h − 1
hm+1− 1 · (h(m+1)n− 1) L(f, Pn) =
n
X
i=1
(ahi−1)mahi−1(h − 1) = am+1(h − 1)
m
X
i=1
h(m+1)(i−1)= am+1 (h − 1)
hm+1− 1(h(m+1)n− 1).
Use the fact that hn= b/a, one finds that
h(m+1)n = (hn)m+1= b a
m+1
= bm+1 am+1. Hence U (f, Pn) and L(f, Pn) can be rewritten as
U (f, Pn) = h − 1
hm+1− 1(bm+1− am+1)hm L(f, Pn) = h − 1
hm+1− 1(bm+1− am+1).
By definition,
(1.2) h − 1
hm+1− 1(bm+1− am+1) ≤ Z b
a
xmdx ≤ Z b
a
xmdx ≤ h − 1
hm+1− 1(bm+1− am+1)hm. Let us recall a formula:
1 + h + · · · + hm= hm+1− 1 h − 1 for any m ≥ 1. Hence
h − 1
hm+1− 1 = 1
1 + h + · · · + hm. Recall that
n→∞lim h = lim
n→∞
n
rb a = 1.
Hence we obtain that
n→∞lim
h − 1
hm+1− 1 = 1
n→∞lim(1 + h + · · · + hm)= 1 m + 1. Suppose that we know the following fact:
Lemma 1.1. Let (an) and (bn) be sequences and lim
n→∞an= a, lim
n→∞bn = b.
Then
n→∞lim anbn= ab.
3
Using the above fact, we have
n→∞lim
h − 1
hm+1− 1(bm+1− am+1)hm= (bm+1− am+1) lim
n→∞
h − 1
hm+1− 1· lim
m→∞hm=bm+1− am+1 m + 1 . By taking n → ∞ of (1.2), we obtain that
bm+1− am+1
m + 1 ≤
Z b a
xmdx ≤ Z b
a
xmdx ≤ bm+1− am+1 m + 1 . This shows that
Z b a
xmdx = Z b
a
xmdx = bm+1− am+1 m + 1 , i.e. f is Riemann integrable and
Z b a
xmdx = bm+1− am+1 m + 1 . Let us define a function F : [a, b] → R by
F (x) = Z x
a
tmdt, a ≤ x ≤ b.
By the previous example, we know that
F (x) = xm+1− am+1
m + 1 , a ≤ x ≤ b.
Let us recall the binomial theorem:
Theorem 1.1. (Binomial Theorem) Let x, y be real numbers (assume nonzero). For any natural number n,
(x + y)n= xn+n 1
xn−1y + · · · +
n n − 1
xyn−1+ yn=
n
X
i=0
n i
xn−iyi.
Heren i
= n!
i!(n − i)!.
(We will define the derivative of a function more precisely later). Let a < x < b and h be a number so that x + h is contained in (a, b). Then
F (x + h) − F (x)
h = xm+ h
m + 1
m + 1 2
xm−1+ · · · + hm
. Hence
F0(x) = lim
h→0
F (x + h) − F (x)
h = xm.
In fact such a statement holds for integration of continuous function. More precisely:
Theorem 1.2. (Fundamental Theorem of Calculus I) Let f : [a, b] → R be a continuous function (It must be Riemann integrable). Define a function F : [a, b] → R by
F (x) = Z x
a
f (t)dt, a ≤ x ≤ b.
Then F is continuous on [a, b] and differentiable on (a, b) so that F0(x) = f (x) for a < x < b.
Theorem 1.3. (Fundamental Theorem of Calculus II) Let f : [a, b] → R be a continuous function.
Suppose G : [a, b] → R is continuous and differentiable on (a, b) so that G0(x) = f (x) for a < x < b.
Then
Z b a
f (x)dx = G(b) − G(a).
