Lesson 25: Evaluating Definite Integrals (Section 041 handout)

V63.0121.041, Calculus I Section 5.3 : Evaluating Definite Integrals December 6, 2010

Section 5.3
Notes
Evaluating Definite Integrals

V63.0121.041, Calculus I

New York University

December 6, 2010

Announcements

Today: Section 5.3
Wednesday: Section 5.4
Monday, December 13: Section 5.5
”Monday,” December 15: ???
Monday, December 20, 12:00–1:50pm: Final Exam

Announcements
Notes

Today: Section 5.3
Wednesday: Section 5.4
Monday, December 13:
Section 5.5
”Monday,” December 15:
???
Monday, December 20,
12:00–1:50pm: Final Exam

V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 2 / 41

Objectives
Notes

Use the Evaluation Theorem
to evaluate definite integrals.
Write antiderivatives as
indefinite integrals.
Interpret definite integrals as
“net change” of a function
over an interval.


1


Outline
Notes

Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
Comparison Properties of the Integral

Examples

The Integral as Total Change

Indefinite Integrals
My first table of integrals

Computing Area with integrals


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 n→∞
i=1

b−a
where ∆x = , and for each i, xi = a + i∆x, and ci is a point in
n
[xi−1 , xi ].

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 and is the same for any choice of ci .
a


Notation/Terminology
Notes

b
f (x) dx
a

— 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


2


Example
1
4 Notes
Estimate dx using the midpoint rule and four divisions.
0 1 + x2

Solution
Dividing up [0, 1] into 4 pieces gives

1 2 3 4
x0 = 0, x1 = , x2 = , x3 = , x4 =
4 4 4 4
So the midpoint rule gives

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
150, 166, 784
= ≈ 3.1468
47, 720, 465


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


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
Theorem
c b c
5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c.
a a b


3


Deﬁnite Integrals We Know So Far
Notes

If the integral computes an
area and we know the area,
we can use that. For
y
instance,
1
π
1 − x 2 dx =
0 2

By brute force we computed x
1 1
1 1
x 2 dx = x 3 dx =
0 3 0 4


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


Estimating an integral with inequalities
Notes
Example
2
1
Estimate dx using Property 8.
1 x

Solution
Since
1 1 1
1 ≤ x ≤ 2 =⇒ ≤ ≤
2 x 1
we have
2
1 1
· (2 − 1) ≤ dx ≤ 1 · (2 − 1)
2 1 x
or
2
1 1
≤ dx ≤ 1
2 1 x


4


Outline
Notes


Examples





Socratic proof
Notes

The deﬁnite integral of
velocity measures
displacement (net distance)
The derivative of
displacement is velocity
So we can compute
displacement with the
deﬁnite integral or the
antiderivative of velocity
But any function can be a
velocity function, so . . .


Theorem of the Day
Notes

Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F for another function F , then
b
f (x) dx = F (b) − F (a).
a


5


Proving the Second FTC
Notes

b−a
Divide up [a, b] into n pieces of equal width ∆x = as usual. For
n
each i, F is continuous on [xi−1 , xi ] and diﬀerentiable on (xi−1 , xi ). So
there is a point ci in (xi−1 , xi ) with

F (xi ) − F (xi−1 )
= F (ci ) = f (ci )
xi − xi−1

Or
f (ci )∆x = F (xi ) − F (xi−1 )


Proof continued
Notes

We have for each i

f (ci )∆x = F (xi ) − F (xi−1 )

Form the Riemann Sum:
n n
Sn = f (ci )∆x = (F (xi ) − F (xi−1 ))
i=1 i=1
= (F (x1 ) − F (x0 )) + (F (x2 ) − F (x1 )) + (F (x3 ) − F (x2 )) + · · ·
· · · + (F (xn−1 ) − F (xn−2 )) + (F (xn ) − F (xn−1 ))
= F (xn ) − F (x0 ) = F (b) − F (a)


Proof Completed
Notes

We have shown for each n,

Sn = F (b) − F (a)

so in the limit
b
f (x) dx = lim Sn = lim (F (b) − F (a)) = F (b) − F (a)
a n→∞ n→∞


6


Computing area with the Second FTC
Notes

Example
Find the area between y = x 3 and the x-axis, between x = 0 and x = 1.

Solution

1 1
x4 1
A= x 3 dx = =
0 4 0 4

Here we use the notation F (x)|b or [F (x)]b to mean F (b) − F (a).
a a


Computing area with the Second FTC
Notes
Example
Find the area enclosed by the parabola y = x 2 and y = 1.

1

−1 1

Solution

1 1
x3 1 1 4
A=2− x 2 dx = 2 − =2− − − =
−1 3 −1 3 3 3


Computing an integral we estimated before
Notes
Example
1
4
Evaluate the integral dx.
0 1 + x2

Solution


7


Computing an integral we estimated before
Notes

Example
2
1
Evaluate dx.
1 x

Solution


Outline
Notes


Examples





Notes

Another way to state this theorem is:
b
F (x) dx = F (b) − F (a),
a

or the integral of a derivative along an interval is the total change between
the sides of that interval. This has many ramiﬁcations:

Theorem
If v (t) represents the velocity of a particle moving rectilinearly, then
t1
v (t) dt = s(t1 ) − s(t0 ).
t0


8


Notes

b
F (x) dx = F (b) − F (a),
a


Theorem
If MC (x) represents the marginal cost of making x units of a product, then
x
C (x) = C (0) + MC (q) dq.
0


Notes

b
F (x) dx = F (b) − F (a),
a


Theorem
If ρ(x) represents the density of a thin rod at a distance of x from its end,
then the mass of the rod up to x is
x
m(x) = ρ(s) ds.
0


Outline
Notes


Examples





9


A new notation for antiderivatives
Notes

To emphasize the relationship between antidiﬀerentiation and integration,
we use the indeﬁnite integral notation

f (x) dx

for any function whose derivative is f (x). Thus

x 2 dx = 1 x 3 + C .
3


Notes

[f (x) + g (x)] dx = f (x) dx + g (x) dx

x n+1
x n dx = + C (n = −1) cf (x) dx = c f (x) dx
n+1
1
e x dx = e x + C dx = ln |x| + C
x
x ax
sin x dx = − cos x + C a dx = +C
ln a

cos x dx = sin x + C csc2 x dx = − cot x + C

sec2 x dx = tan x + C csc x cot x dx = − csc x + C
1
sec x tan x dx = sec x + C √ dx = arcsin x + C
1 − x2
1
dx = arctan x + C
1 + x2


Outline
Notes


Examples





10


Example
Notes
Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the
vertical lines x = 0 and x = 3.

Solution


Graph
Notes


Interpretation of “negative area” in motion
Notes


11


What about the constant?
Notes

It seems we forgot about the +C when we say for instance
1 1
x4 1 1
x 3 dx = = −0=
0 4 0 4 4

But notice
1
x4 1 1 1
+C = +C − (0 + C ) = +C −C =
4 0 4 4 4

no matter what C is.
So in antidifferentiation for definite integrals, the constant is
immaterial.


Summary
Notes

The second Fundamental Theorem of Calculus:
b
f (x) dx = F (b) − F (a)
a

where F = f .
Definite integrals represent net change of a function over an interval.
We write antiderivatives as indefinite integrals f (x) dx


Notes

12

Lesson 25: Evaluating Definite Integrals (Section 041 handout)

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