Lesson 23: The Definite Integral (handout)

V63.0121.006/016, Calculus I Section 5.2 : The Definite Integral April 15, 2010

Section 5.2
Notes
The Definite Integral
V63.0121.006/016, Calculus I

New York University

April 15, 2010

Announcements

April 16: Quiz 4 on §§4.1–4.4
April 29: Movie Day!!
April 30: Quiz 5 on §§5.1–5.4
Monday, May 10, 12:00noon (not 10:00am as previously announced)
Final Exam

Announcements
Notes

April 16: Quiz 4 on
§§4.1–4.4
April 29: Movie Day!!
April 30: Quiz 5 on
§§5.1–5.4
Monday, May 10, 12:00noon
(not 10:00am as previously
announced) Final Exam

V63.0121.006/016, Calculus I (NYU) Section 5.2 The Definite Integral April 15, 2010 2 / 28

Objectives
Notes

Compute the definite
integral using a limit of
Riemann sums
Estimate the definite
integral using a Riemann
sum (e.g., Midpoint Rule)
Reason with the definite
integral using its elementary
properties.


1


Outline
Notes

Recall

The definite integral as a limit

Estimating the Definite Integral

Properties of the integral

Comparison Properties of the Integral


Cavalieri’s method in general
Notes
Let f be a positive function defined on the interval [a, b]. We want to find
the area between x = a, x = b, y = 0, and y = f (x).
For each positive integer n, divide up the interval into n pieces. Then
b−a
∆x = . For each i between 1 and n, let xi be the ith step between a
n
and b. So

x0 = a
b−a
x1 = x0 + ∆x = a +
n
b−a
x2 = x1 + ∆x = a + 2 · ...
n
b−a
xi = a + i · ...
n
b−a
xn = a + n · =b
x n
x0 x1 . . . xi . . .xn−1xn


Forming Riemann sums
Notes
We have many choices of representative points to approximate the area in
each subinterval.

. . . even random points!

x
In general, choose ci to be a point in the ith interval [xi−1 , xi ]. Form the
Riemann sum
n
Sn = f (c1 )∆x + f (c2 )∆x + · · · + f (cn )∆x = f (ci )∆x
i=1


2


Theorem of the (previous) Day
Notes

Theorem

If f is a continuous function on [a, b]
M15 = 7.49968
or has finitely many jump
discontinuities, then
n
lim Sn = lim f (ci )∆x
n→∞ n→∞
i=1

exists and is the same value no
matter what choice of ci we make. x


Outline
Notes

Recall






Notes

Definition
If f is a function defined on [a, b], the definite integral of f from a to b
is the number
b n
f (x) dx = lim f (ci ) ∆x
a ∆x→0
i=1


3


Notation/Terminology
Notes

b n
f (x) dx = lim f (ci ) ∆x
a ∆x→0
i=1

— integral sign (swoopy S)
f (x) — integrand
a and b — limits of integration (a is the lower limit and b the
upper limit)
dx — ??? (a parenthesis? an infinitesimal? a variable?)
The process of computing an integral is called integration or
quadrature


The limit can be simplified
Notes
Theorem
If f is continuous on [a, b] or if f has only finitely many jump
discontinuities, then f is integrable on [a, b]; that is, the definite integral
b
f (x) dx exists.
a

Theorem
If f is integrable on [a, b] then
b n
f (x) dx = lim f (xi )∆x,
a n→∞
i=1

where
b−a
∆x = and xi = a + i ∆x
n


Example: Integral of x
Notes
Example
3
Find x dx
0

Solution
3 3i
For any n we have ∆x = and xi = . So
n n
n n n
3i 3 9
Rn = f (xi ) ∆x = = i
n n n2
i=1 i=1 i=1
9 n(n + 1) 9
= · −→
n2 2 2
3
9
So x dx = = 4.5
0 2


4


Example: Integral of x 2
Notes
Example
3
Find x 2 dx
0

Solution
3 3i
n n
n n 2 n
3i 3 27
Rn = f (xi ) ∆x = = i2
n n n3
i=1 i=1 i=1
27 n(n + 1)(2n + 1) 27
= · −→ =9
n3 6 3
3
So x 2 dx = 9
0


Example: Integral of x 3
Notes
Example
3
Find x 3 dx
0

Solution
3 3i
n n
n n 3 n
3i 3 81
Rn = f (xi ) ∆x = = i3
n n n4
i=1 i=1 i=1
81 n2 (n + 1)2 81
= · −→
n4 4 4
3
81
So x 3 dx = = 20.25
0 4


Outline
Notes

Recall






5


Notes
Example
1
4
Estimate dx using M4 .
0 1 + x2

Solution
1 1 3
We have x0 = 0, x1 = , x2 = , x3 = , x4 = 1.
4 2 4
1 3 5 7
So c1 = , c2 = , c3 = , c4 = .
8 8 8 8
1 4 4 4 4
M4 = + + +
4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2
1 4 4 4 4
= + + +
4 65/64 73/64 89/64 113/64
64 64 64 64
= + + + ≈ 3.1468
65 73 89 113


Estimating the Deﬁnite Integral (Continued)
Notes
Example
1
4
Estimate dx using L4 and R4
0 1 + x2

Answer

4 1 4 4 4
L4 = + + +
1 + (0)2 1 + (1/4)2 1 + (1/2)2 1 + (3/4)2
4
16 4 16
=1+ + + ≈ 3.38118
17 5 25
1 4 4 4 4
R4 = + + +
4 1 + (1/4)2 1 + (1/2)2 1 + (3/4)2 1 + (1)2
16 4 16 1
= + + + ≈ 2.88118
17 5 25 2


Outline
Notes

Recall






6


Notes

Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a, b] and c a constant. Then
b
1. c dx = c(b − a)
a
b b b
2. [f (x) + g (x)] dx = f (x) dx + g (x) dx.
a a a
b b
3. cf (x) dx = c f (x) dx.
a a
b b b
4. [f (x) − g (x)] dx = f (x) dx − g (x) dx.
a a a


Proofs
Notes

Proofs.

When integrating a constant function c, each Riemann sum equals
c(b − a).
A Riemann sum for f + g equals a Riemann sum for f plus a
Riemann sum for g . Using the sum rule for limits, the integral of a
sum is the sum of the integrals.
Ditto for constant multiples
Ditto for diﬀerences


Example Notes
3
3 2
Find x − 4.5x + 5.5x + 1 dx
0

Solution

3
(x 3 −4.5x 2 + 5.5x + 1) dx
0
3 3 3 3
= x 3 dx − 4.5 x 2 dx + 5.5 x dx + 1 dx
0 0 0 0

= 20.25 − 4.5 · 9 + 5.5 · 4.5 + 3 · 1 = 7.5

(This is the function we were estimating the integral of before)


7


Theorem of the (previous) Day
Notes

Theorem

If f is a continuous function on [a, b]
M15 = 7.49968
or has ﬁnitely many jump
discontinuities, then
n
lim Sn = lim f (ci )∆x
n→∞ n→∞
i=1

exists and is the same value no
matter what choice of ci we make. x


More Properties of the Integral
Notes

Conventions:
a b
f (x) dx = − f (x) dx
b a
a
f (x) dx = 0
a
This allows us to have
c b c
5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c.
a a b


Example Notes
Suppose f and g are functions with
4
f (x) dx = 4
0
5
f (x) dx = 7
0
5
g (x) dx = 3.
0
Find
5
(a) [2f (x) − g (x)] dx
0
5
(b) f (x) dx.
4


8


Solution Notes
We have
(a)
5 5 5
[2f (x) − g (x)] dx = 2 f (x) dx − g (x) dx
0 0 0
= 2 · 7 − 3 = 11

(b)
5 5 4
f (x) dx = f (x) dx − f (x) dx
4 0 0
=7−4=3


Outline
Notes

Recall






Notes
Theorem
Let f and g be integrable functions on [a, b].
6. If f (x) ≥ 0 for all x in [a, b], then
b
f (x) dx ≥ 0
a

7. If f (x) ≥ g (x) for all x in [a, b], then
b b
f (x) dx ≥ g (x) dx
a a

8. If m ≤ f (x) ≤ M for all x in [a, b], then
b
m(b − a) ≤ f (x) dx ≤ M(b − a)
a


9


Notes
Example
2
1
Estimate dx using the comparison properties.
1 x

Solution
Since
1 1
≤x ≤
2 1
for all x in [1, 2], we have
2
1 1
·1≤ dx ≤ 1 · 1
2 1 x


Notes

Notes

10

Lesson 23: The Definite Integral (handout)

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