Download Lecture Notes 31: Riemann Sums, Definite Integral, Properties of

Lecture Notes 31: Riemann Sums, Definite
Integral, Properties of Definite Integral
Instructor: Anatoliy Swishchuk
Department of Mathematics & Statistics
University of Calgary, Calgary, AB, Canada
MATH 265 ’University Calculus I’
L01 Winter 2017
Outline of Lecture
1. Short Intro
2. Partitions and Riemann Sums
3. The Definite Integral
4. General Riemann Sums
5. Examples
Short Intro
In this lecture we generalize and make more precise the procedure
used for finding areas developed in our previous Lecture 30.
We use this procedure to define the definite integral of a function
f on an interval I.
Short Intro
We assume that f (x) is defined and continuous on the closed,
finite interval [a, b].
We no longer assume that the values of f are nonnegative.
Partitions and Riemann Sums
Let
P := {x0, x1, , x2, ..., xn−1, xn},
a = x0 < x1 < x2... < xn−1 < xn = b.
Such a set P is called a partition of [a, b]; it divides [a, b] into n
subintervals of which the ith is [xi−1, xi].
We call these the subintervals of the partition P . Their lengths
are ∆xi = xi − xi−1, 1 ≤ i ≤ n.
We also define the norm of the partition P and denote it ||P || :
||P || = max1≤i≤n ∆xi.
Partitions and Riemann Sums (cont’d)
Since f is continuous on each subinterval [xi−1, xi] of P, it takes
on maximum and minimum values at points of that interval.
Thus there are numbers li and ui in [xi−1, xi] such that
f (li) ≤ f (x) ≤ f (ui),
xi−1 ≤ x ≤ xi.
If f (x) ≥ 0 on [a, b], then f (li)∆xi and f (ui)∆xi represent the
areas of rectangles having the interval [xi−1, xi] on the x-axis as
base, and having tops passing through the lowest and highest
points, respectively, on the graph of f on that interval.
Partitions and Riemann Sums (cont’d)
If Ai is that part of the area under y = f (x) and above the x-axis
that lies in the interval strip between x = xi−1 and x = xi, then
f (li)∆xi ≤ Ai ≤ f (ui)∆xi.
If f can have negative values, then one or both of f (li)∆xi and
f (ui)∆xi can be negative and will then represent the negative of
the area of a rectangle lying below the x-axis. In any event, we
always have
f (li)∆xi ≤ f (ui)∆xi.
Partitions and Riemann Sums (cont’d)
Def. 1 (Upper and Lower Riemann Sums)
The lower (Riemann) sum, L(f, P ), and the upper (Riemann)
sum, U (f, P ), for the function f and the partition P are defined
by:
L(f, P ) =
n
X
f (li)∆xi,
i=1
U (f, P ) =
n
X
i=1
f (ui)∆xi.
Example 1
Calculate the lower and upper Riemann sums for the function
f (x) = x2 on the interval [0, a] (where a > 0), corresponding the
partition Pn of [0, a] into n subintervals of equal length.
Sol. Each subinterval of Pn has length ∆x = a/n, and the
division points are given by xi = ia/n for i = 0, 1, 2, ..., n. Since
x2 is increasing of [0, a], its minimum and maximum values over
the ith subinterval [xi−1, xi] occur at li = xi−1 and ui = xi,
respectively.
Example 1 (cont’d)
Thus, the lower Riemann sum of f for Pn is
3 P
Pn
a
n
2
2
L(f, Pn) =
i=1 (xi−1 ) ∆x = n3 i=1 (i − 1)
3 Pn−1
3 (n−1)n(2(n−1)+1)
(n−1)(2n−1)a3
a
a
2
= n3 j=1 j = n3
=
,
6
6n2
see Appendix, Lecture 26. Similarly, the upper Riemann sum is:
Pn
2 ∆x = a3 Pn (i)2
(x
)
i=1 i
n3 i=1
3
a3 Pn
2 = a3 (n)(n+1)(2n+1) = (n+1)(2n+1)a .
i
= n
3
i=1
6
n3
6n2
U (f, Pn) =
The Definite Integral
Def 2. Suppose there is exactly one number I such that for
every partition P of [a, b] we have
L(f, P ) ≤ I ≤ U (f, P ).
Then we say that the function f is integrable on [a, b], and we
call I the definite integral of f on [a, b].
The definite integral is denoted by the symbol
I=
Z b
a
f (x)dx.
Important Remarks
The definite integral of f (x) is a number, it is not a function of
x. It depends on a and b and on particular function f , but not on
variable x. This variable x is a dummy variable like the variable
Pn
i in the sum i=1 f (i).
Replacing x with another variable does not change the value of
the integral:
Z b
a
f (x)dx =
Z b
a
f (t)dt.
Rb
Names for Various Parts of a f (x)dx
i)
R
is called the integral sign;
ii) a and b are called the limits of integration; a is the lower
limit, b is the upper limit;
iii) the function f is the integrand; x is the variable of integration;
iv) dx is the differential of x. It replaces ∆x in the Riemann
sums.
Eample 2
Show that f (x) = x2 is integrable over the interval [0, a], where
R
a > 0, and evaluate 0a x2dx.
Sol. Let us evaluate the limits as n → +∞ of the lower and
upper sums of f over [0, a] obtained in Example 1 above:
3
(n−1)(2n−1)a3
a
= 3,
limn→+∞ L(f, Pn) = limn→+∞
6n2
(n+1)(2n+1)a3
a3 .
limn→+∞ U (f, Pn) = limn→+∞
=
3
6n2
If L(f, Pn) ≤ I ≤ U (f, Pn), we must have I = a3/3. Thus, f (x) =
x2 is integrable over [0, a], and
Z a
0
f (x)dx =
Z a
0
a3
2
x dx =
.
3
Important Remark
For all partitions P on [0, a], we have
L(f, P ) ≤
Z b
a
f (x)dx ≤ U (f, P ).
Rb
In general, a f (x)dx is the area of that part of R lying above the
x-axis (where f (x) ≥ 0) minus the area of the part lying below
the x-axis (where f (x) ≤ 0).
General Riemann Sums
Let
P := {x0, x1, , x2, ..., xn−1, xn},
a = x0 < x1 < x2... < xn−1 < xn = b,
be a partition of [a, b] and let ||P || = max1≤i≤n ∆xi. In each
subinterval [xi−1, xi] of P pick a point ci (called a tag). Let
c = (c1, c2, ..., cn) denote the set of these tags. The sum
R(f, P, c) =
n
X
f (ci)∆xi
i=1
is called the Riemann sum of f on [a, b] corresponding to partition P and tags c.
General Riemann Sums (cont’d)
For any choice of c, the Riemann sum R(f, P, c) satisfies
L(f, P ) ≤ R(f, P, c) ≤ U (f, P ).
Also,
lim
n→+∞;||P ||→0
R(f, P, c) =
Z b
a
f (x)dx.
Integrability of a Function f on [a, b]
Theorem 1. If f is continuous on [a, b], then f is integrable on
[a, b].
Example 3
Express the limit limn→+∞
tegral.
Pn
2 (1 + 2i−1 )1/3 as a definite ini=1 n
n
Sol. The factor 2/n suggests that the interval of integration
has length 2 and is partitioned into equal subintervals, each of
length 2/n. Let ci = (2i − 1)/n for i = 1, 2, ..., n. When n → +∞,
then c1 = 1/n → 0 and cn = (2n − 1)/n → 2. Thus, the interval is
[0, 2] and the points of the partition are xi = 2i/n. Observe that
xi−1 = (2i − 2)/n < ci < 2i/n = xi for each i, so that the sum is
indeed a Riemann sum for f (x) over [0, 2]. Since f is continuous
on that interval, it is integrable there, and
n
X
2
2
2i − 1 1/3
(1 +
)
=
(1 + x)1/3dx.
lim
n→+∞ i=1 n
n
0
Z
References
1) Calculus: Early Transcendental, 2016, An Open Text, by
David Guichard: https : //lalg1.lyryx.com/textbooks/CALCU LU S
1/ucalgary/winter2016/math265/Guichard
− Calculus − EarlyT rans − U of Calgary − M AT H265 − W 16.pdf
2) Optional Textbook: Essential Calculus, Early Transcendental,
2013, by J. Stewart, 2nd edition, Brooks/Cole