Math 121,
Lecture 1
Lior Silberman
The
Construction
Properties of
the definite
integral
Math 121: Honours Integral Calculus
Lecture 6
Lior Silberman
January 11th , 2012
Last time
Math 121,
Lecture 1
Lior Silberman
f : [a, b] → R bounded.
The
Construction
P : a = x0 < x1 < · · · < xn = b a paritition.
Properties of
the definite
integral
Spacings ∆xi = xi − xi−1 .
Mesh: δ (P) = max {∆xi }ni=1 (longest spacing).
mi =
inf
x ∈[xi−1 ,xi ]
f (x ), Mi =
Riemann sums L(f ; P) =
U(f ; P) = ∑ni=1 Mi ∆xi .
Definition
sup
f (x ).
x ∈[xi−1 ,xi ]
∑ni=1 mi ∆xi ,
b
Say f is integrable on [a, b] and that a f (x )dx = I if I ∈ Ris
the unique number so that L(f ; P) ≤ I ≤ U(f ; P) for all P.
Examples
Math 121,
Lecture 1
Lior Silberman
The
Construction
Properties of
the definite
integral
b
cdx = c(b − a).
(
1 x rational
Dirichlet’s function D(x ) =
is not
0 x irrational
integrable on any interval (mi = 0, Mi = 1).
(
D(x ) 0 ≤ x ≤ 12
What if f (x ) =
on [0, 1]?
1
1
2 ≤x ≤1
(
1 x =0
What if f (x ) =
?
0 0<x ≤1
f constant:
a
The choice of partition
Math 121,
Lecture 1
Lior Silberman
Theorem
The following are equivalent:
The
Construction
1
I is the unique number between the lower and upper sums;
Properties of
the definite
integral
2
I = limδ (P)→0 L(f ; P) = limδ (P)→0 U(f ; P);
3
The sums for the uniform partition converge to I.
Why non-uniform partitions?
Example
f (x ) = log x on [a, b] (a > 0). The natural partition is
i/n
xi = a ba
. After complicated calculation (see notes) get
b
log dx = (b log b − b) − (a log a − a) .
a
The interval
Math 121,
Lecture 1
Theorem
Lior Silberman
The
Construction
Let f be Riemann integrable on [a, b]. Then f is integrable on
any sub-interval.
Properties of
the definite
integral
Theorem
Let f be integrable on [a, b], [b, c]. Then f is integrable on
[a, c] and
b
f (x )dx +
a
Proof.
Concatenate partitions.
c
f (x )dx =
b
c
f (x )dx .
a
The interval
Math 121,
Lecture 1
Lior Silberman
The
Construction
Properties of
the definite
integral
Definition
If b < a set
to zero.
a
b
f (x )dx = −
b
a
f (x )dx . If a = b set the integral
Example
(
5
2≤x <3
f (x ) =
. Then
−2 3 < x ≤ 5
5
2
f (x )dx =
Example
3
2
f (x )dx +
b
5
3
f (x )dx = 5 · 1 + (−2) · 2 = 1.
2
We showed that 0 xdx = b2 (right triangle with both sides of
b a b
length b). Since 0 = 0 + a it follows that
b
b
a
b2
a2
a xdx = 0 xdx − 0 xdx = 2 − 2 .
The function
Math 121,
Lecture 1
Lior Silberman
The
Construction
Properties of
the definite
integral
Theorem
Let f , g be Riemann integrable on [a, b]. Let A, B ∈ R. Then
Af + Bg is integrable on [a, b] and
b
a
a
b
a
(Ax + B)dx = A
b
a
xdx + B
b
a
b
f (x )dx + B
(Af (x ) + Bg(x )) dx = A
Example
b
1dx = A
g(x )dx .
a
b2
2
2
− a2
+ B(b − a).
The function
Math 121,
Lecture 1
Lior Silberman
The
Construction
Properties of
the definite
integral
Theorem
Let f be continous on [a, b]. Then f is integrable on [a, b].
Proof.
Continuity: f does not fluctuate much on small intervals. But
this means that if δ (P) is small then U(f ; P) − L(f ; P) is small
(at most b − a times the maximal fluctuation). It follows that
there is at most one real number between all the lower and
upper sums.