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Mathematics

MATH 118

Steve Spencer

Winter

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Chapter 1. Review of the Riemann Integral
The Riemann Integral
1.1 De▯nition: A partition of the closed interval [a;b] is a set X = fx ;x ;▯▯▯;x g
0 1 n
with
a = x < x < x < ▯▯▯ < x = b:
0 1 2 n
The intervals [xi▯1;xi] are called the subintervals of [a;b], and we write
▯ i = x i x i▯1
th
for the size of the i subinterval. Note that
Xn
▯ i = b ▯ a:
i=1
The size of the partition X, denoted by jXj is
▯ ▯
jXj = max ▯ x 1 ▯ i ▯ n :
i
1.2 De▯nition: Let X be a partition of [a;b], and let f : [a;b] ! R be bounded. A
Riemann sum for f on X is a sum of the form
Xn
S = f(t )▯ x for some t 2 [x ;x ]:
i i i i▯1 i
i=1
The points tiare called sample points.
1.3 De▯nition: Let f : [a;b] ! R be bounded. We say that f is (Riemann) integrable
on [a;b] when there exists a number I with the property that for every ▯ > 0 there exists
▯ > 0 such that for every partition X of [a;b] with jXj < ▯ we have jS ▯ Ij < ▯ for every
Riemann sum for f on X, that is
▯n ▯
▯ ▯
▯ f(ti)▯ i ▯ I ▯ ▯:
▯i=1 ▯
for every choice of i 2 [i▯1;x i The number I can be shown to be unique. It is called the
(Riemann) integral of f on [a;b], and we write
Z Z
b b
I = f , or I = f(x)dx:
a a
1.4 Example: Show that the constant function f(x) = c is integrable on any interval
Z b
[a;b] and we have c dx = c(b ▯ a).
a
Solution: The solution is left as an exercise.
1 1.5 Example: Show that the identity function f(x) = x is integrable on any interval
Z
b 1 2 2
[a;b], and we have x dx = 2(b ▯ a ).
a
Solution: Let ▯ > 0. Choose ▯ = 2▯ . Let X be any partition of [a;b] with jXj < ▯. Let
n b▯a n
P P 1 2 2
ti2 [xi▯1 ;xi] and set S = f(ti)▯ix = ti▯ i. We must show that jS▯ 2 (b ▯a )j < ▯.
i=1 i=1
Notice that
n n
Pn X X 2 2
(xi+ x i▯1)▯ i = (xi+ x i▯1)(xi▯ x i▯1) = xi ▯ xi▯1
i=1 i=1 i=1
= (x12 ▯ x 0 + (x 22▯ x 1 + ▯▯▯ + (x n▯12 ▯ xn▯2 2) + (xn2 ▯ x n▯12)
= ▯x 02 + (x1 2▯ x 1 + ▯▯▯ + (x n▯1 2 ▯ xn▯1 2) + xn2
= x n2▯ x 02 = b ▯ a 2
▯ 1 ▯ 1 1
and that when t i [x i▯1;x i we have t i 2(xi+ x i▯1 ) ▯ 2(xi▯ x i▯1) = 2 x,iand so
▯ ▯
▯ 1 2 2 ▯ ▯ P 1Pn ▯
S ▯ 2(b ▯ a ) = ▯ ti▯ix ▯ 2 (xi+ x i▯1 )▯ix▯
i=1 i=1
▯ P ▯ ▯ ▯
= ▯ ti▯ (x +ix i+1) ▯ i ▯
i=1 2
P ▯ 1 ▯
▯ ti▯ (2 + i i+1) ▯ix
i=1
P P
▯ 1▯ i▯ xi▯ 1▯▯ i
i=1 2 i=1 2
= ▯(b ▯ a) = ▯:
2
▯
1.6 Example: Let f(x) = 1 if x 2 Q Show that f is not integrable on [0;1].
0 if x = Q:
R 1
Solution: Suppose, for a contradiction, that f is integrable on [0;1], and write I = 0 f.
Let ▯ = . Choose ▯ so that for every partition X with jXj < ▯ we have jS▯Ij < 1for every
2 n2
P
Riemann sum S for f on X. Choose a partition X with jXj < ▯. Let S = 1 f(ti)▯ i
i=1
Pn
where each t i [x i▯1 ;xi] is chosen with ti2 Q, and let S =2 f(si)▯ i where each
i=1
si2 [x i▯1;x i is chosen with si2= Q. Note that we have jS ▯ 1j < 1 and jS 2 Ij < 1.
n n 2 2
P P
Since each ti2 Q we have f(t ) i 1 and so S = 1 f(ti)▯ix = ▯ i = 1 ▯ 0 = 1, and
in1 i=1
P 1
since each si2= Q we have f(s )i= 0 and so S =2 f(si)▯ i = 0. Since jS 1 Ij < 2 we
i=1
have j1 ▯ Ij < 1 and so 1 < I < 3, and since jS2▯ Ij < 1 we have j0 ▯ Ij < 1 and so
1 1 2 2 2 2 2
▯ 2 < I < ,2giving a contradiction.
2 Integrals of Continuous Functions
1.7 Theorem: (Continuous Functions are Integrable) Let f : [a;b] ! R be continuous.
Then f is integrable on [a;b].
Proof: We omit the proof, which is quite di▯cult.
1.8 Note: Let f be integrable on [a;b]. Lnt X be any sequence of partitions of [a;b] with
n!1m jXnj = 0. Let n be any Riemann sum for f on Xn. Then fSng converges with
Z b
lim S = f(x)dx:
n!1 n a
R b
Proof: Write I = a f. Given ▯ > 0, choose ▯ > 0 so that for every partition X of [a;b]
with jXj < ▯ we have jS ▯ Ij < ▯ for every Riemann sum S for f on X, and then choose
N so that n > N =) jX n < ▯. Then we have n > N =) jS n Ij < ▯.
1.9 Note: Let f be integrable on [a;b]. If we let n be the partition of [a;b] into n
equal-sized subintervals, and we net S be the Riemann sum on X using right-endpoints,
then by the above note we obtain the formula
Z n
b X b▯a b▯a
f(x)dx = lim f(x n;i▯n;i , where xn;i a + n i and ▯n;i = n :
a n!1 i=1
Z 2
1.10 Example: Find 2 dx.
0
Solution: Let f(x) = 2 . Note that f is continuous and hence integrable, so we have
Z 2 Xn Xn ▯ ▯▯ ▯ Xn ▯ ▯
2 dx = lim f(x )▯ x = lim f 2i 2 = lim 22i=n 2
n!1 n;i n;i n!1 n n n!1 n
0 i=1 i=1 i=1
1=n
= lim 2 ▯ 4 ▯ 4 ▯ 1 , by the formula for the sum of a geometric sequence
n!1 n 41=n▯ 1
1
▯ 1=n▯▯ 1 ▯ n x
= lim 6 ▯ 4 lim ▯ 1=n ▯ = 6 lim 1=n = 6 lim x
n!1 n!1 n 4 ▯ 1 n!1 4 ▯ 1 x!0 4 ▯ 1
= 6 lim 1 , by l’opital’s Rule
x!0 ln4 ▯ 4
6 3
= ln 4= ln 2
3 1.11 Lemma: (Summation Formulas) We have
X X Xn Xn 2 2
n(n + 1) 2 n(n + 1)(2n + 1) 3 n (n + 1)
1 = n ; i = 2 ; i = 6 ; i = 4
i=1 i=1 i=1 i=1
Proof: These formulas could be proven by induction, but we give a more constructive
P P P ▯2 2▯
proof. It is obvious tha1 = 1+1+▯▯▯1 = n. To ▯nd i, consider i ▯(i▯1) .
i=1 i=1 n=1
On the one hand, we have
P ▯ 2 2▯ 2 2 2 2 2 2 2 2
i ▯ (i ▯ 1) = (1 ▯ 0 ) + (2 ▯ 1 ) + ▯▯▯ + ((n ▯ 1) ▯ (n ▯ 2) ) + (n ▯ (n ▯ 1) )
i=1
2 2 2 2 2 2 2 2
= ▯0 + (1 ▯ 1 ) + (2 ▯ 2 ) + ▯▯▯ + ((n ▯ 1) ▯ (n ▯ 1) ) + n
2
= n
and on the other hand,
n n n n n
P ▯ 2 2▯ P ▯ 2 2 ▯ P P P
i ▯ (i ▯ 1) = i ▯ (i ▯ 2i + 1) = (2i ▯ 1) = 2 i ▯ 1
i=1 i=1 i=1 i=1 i=1
Pn Pn
Equating these gives n = 2 i ▯ 1 and so
i=1 i=1
Pn P
2 i = n + 1 = n + n = n(n + 1);
i=1 i=1
P 2 P ▯3 3▯
as required. Next, to ▯ndi , consider i ▯ (i ▯ 1) . On the one hand we have
n=1 i=1
n
P ▯3 3▯ 3 3 3 3 3 3 3 3
i ▯ (i ▯ 1)= (1 ▯ 0 ) + (2 ▯ 1 ) + (3 ▯ 2 ) + ▯▯▯ + (n ▯ (n ▯ 1) )
i=1
= ▯0 + (1 ▯ 1 ) + (2 ▯ 2 ) + ▯▯▯ + ((n ▯ 1) ▯ (n ▯ 1) ) + n
= n 3
and on the other hand,
Pn ▯ ▯ P ▯ ▯
i ▯ (i ▯ 1)= i ▯ (i ▯ 3i + 3i ▯ 1)
i=1 i=1
P 2 Pn 2 Pn Pn
= (3i ▯ 3i + 1) = 3 i ▯ 3 i + 1:
i=1 i=1 i=1 i=1
n n n
Equating these gives n = 3 i ▯ 3P i +P 1 and so
i=1 i=1 i=1
P Pn P
6 i = 2n + 6 i ▯ 2 1 = 2n + 3n(n + 1) ▯ 2n = n(n + 1)(2n + 1)
i=1 i=1 i=1
Pn 3 Pn ▯4 4▯
as required. Finally, to ▯nd , consider i ▯ (i ▯ 1) . On the one hand we have
i=1 i=1
n
P ▯4 4▯ 4
i ▯ (i ▯ 1) = n ;
i=1

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