Lesson 24: Evaluating Definite Integrals (handout)

V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010

Section 5.3
Notes
Evaluating Definite Integrals

V63.0121.006/016, Calculus I

New York University

April 20, 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
Image credit: docman

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 5.3 Evaluating Definite Integrals April 20, 2010 2 / 48

Homework: The Good
Notes

Most got problems 1 and 3 right.


1


Homework: The Bad (steel pipe)
Notes
Problem
A steel pipe is being carried down a hallway 9 ft wide. At the end of the
hall there is a right-aangled turn into a narrower hallway 6 ft wide. What is
the length of the longest pipe that can be carried horizontally around the
corner?

θ

θ 6 6
6 sec
9

θ θ
9 csc

9


Solution
Notes
Solution
The longest pipe that barely fits is the smallest pipe that almost doesn’t
fit. We want to find the minimum value of

f (θ) = a sec θ + b csc θ

on the interval 0 < θ < π/2. (a = 9 and b = 6 in our problem.)

f (θ) = a sec θ tan θ − b csc θ cot θ
sin θ cos θ a sin3 θ − b cos3 θ
=a −b 2 =
cos2 θ sin θ sin2 θ cos2 θ
So the critical point is when
b
a sin3 θ = b cos3 θ =⇒ tan3 θ =
a


Finding the minimum
Notes
If f (θ) = a sec θ tan θ − b csc θ cot θ, then

f (θ) = a sec θ tan2 θ + a sec3 θ + b csc θ cot2 θ + b csc3 θ

which is positive on 0 < θ < π/2.
So the minimum value is

f (θmin ) = a sec θmin + b csc θmin
1/3
b b
where tan3 θmin = =⇒ tan θmin = .
a a
Using
1 + tan2 θ = sec2 θ 1 + cot2 θ = csc2 θ
We get the minimum value is

2/3
b a 2/3
min = a 1+ +b 1+
a b

2


Simplifying
Notes

2/3
b a 2/3
min = a 1+ +b 1+
a b

b 2/3 a2/3 a2/3 b 2/3
=b + +a + 2/3
b 2/3 b 2/3 a2/3 a
b a
= b 2/3 + a2/3 + 1/3 a2/3 + b 2/3
b 1/3 a
= b 2/3 b 2/3 + a2/3 + a2/3 a2/3 + b 2/3
= (b 2/3 + a2/3 ) b 2/3 + a2/3
2/3 2/3 3/2
= (a +b )

If a = 9 and b = 6, then min ≈ 21.070.


Homework: The Bad (Diving Board)
Notes

Problem
If a diver of mass m stands at the end of a diving board with length L and
linear density ρ, then the board takes on the shape of a curve y = f (x),
where
EIy = mg (L − x) + 2 ρg (L − x)2
1

E and I are positive constants that depend of the material of the board
and g < 0 is the acceleration due to gravity.
(a) Find an expression for the shape of the curve.
(b) Use f (L) to estimate the distance below the horizontal at the end of
the board.


Notes
Solution
We have
EIy (x) = mg (L − x) + 1 ρg (L − x)2
2

Antidiﬀerentiating once gives

EIy (x) = − 1 mg (L − x)2 − 1 ρg (L − x)3 + C
2 6

Once more:

EIy (x) = 1 mg (L − x)3 +
6
1
24 ρg (L − x)4 + Cx + D

where C and D are constants.


3


Don’t stop there!
Notes
Plugging y (0) = 0 into

EIy (x) = 1 mg (L − x)3 +
6
1
24 ρg (L − x)4 + Cx + D

gives
1 1
0 = 1 mgL3 +
6
1
24 ρgL
4
+ D =⇒ D = − mgL3 − ρgL4
6 24
Plugging y (0) = 0 into

EIy (x) = − 1 mg (L − x)2 − 1 ρg (L − x)3 + C
2 6

gives
1 1
0 = − 1 mgL2 − 6 ρgL3 + C =⇒ C = mgL2 + ρgL3
2
1
2 6


Solution completed
Notes
So

EIy (x) = 1 mg (L − x)3 +
6
1
24 ρg (L − x)4
1 1 1 1
+ mgL2 + L3 x − mgL3 − ρgL4
2 6 6 24

which means
1 1 1 1
EIy (L) = mgL2 + L3 L − mgL3 − ρgL4
2 6 6 24
1 3 1 4 1 1
= mg L + L − mgL3 − ρgL4
2 6 6 24
1 1
= mgL3 + ρgL4
3 8
gL3 m ρL
=⇒ y (L) = +
EI 3 8


Homework: The Ugly
Notes

Some students have gotten their hands on a solution manual and are
copying answers word for word.
This is very easy to catch: the graders are following the same solution
manual.
This is not very productive: the best you will do is ace 10% of your
course grade.
This is a violation of academic integrity. I do not take it lightly.


4


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.


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

The Theorem of the Day
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


5


Notation/Terminology
Notes

b n
f (x) dx = lim f (ci ) ∆x
a n→∞
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


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


6


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 4

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


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


7


Example
2 Notes
1
Estimate dx using Property 8.
1 x

Solution
Since
1 1 1
1 ≤ x ≤ 2 =⇒ ≤ ≤
2 x 1
y
we have
2
1 1
· (2 − 1) ≤ dx ≤ 1 · (2 − 1)
2 1 x
or
1 2
1 x
≤ dx ≤ 1
2 1 x
(Not a very good estimate)


Outline
Notes

Examples





Socratic dialogue
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 an
antiderivative of velocity
But any function can be a
velocity function, so . . .


8


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

Note
In Section 5.3, this theorem is called “The Evaluation Theorem”. Nobody
else in the world calls it that.


Proving the Second FTC
Notes

Proof.
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 )

See if you can spot the invocation of the Mean Value Theorem!


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)


9


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→∞


Verifying earlier computations
Notes
Example

Find the area between y = x 3 the
x-axis, 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

Verifying Archimedes
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


10


Computing exactly what we earlier estimated
Notes
Example
1
4
Evaluate the integral dx.
0 1 + x2

Solution

1 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0
= 4 (arctan 1 − arctan 0)
π
=4 −0 =π
4


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
Example
1
4
Evaluate the integral dx.
0 1 + x2

Solution

1 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0
= 4 (arctan 1 − arctan 0)
π
=4 −0 =π
4


11


Notes

Example
2
1
Evaluate dx.
1 x

Solution

2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1
= ln 2


Example
2 Notes
1
Estimate dx using Property 8.
1 x

Solution
Since
1 1 1
1 ≤ x ≤ 2 =⇒ ≤ ≤
2 x 1
y
we have
2
1 1
· (2 − 1) ≤ dx ≤ 1 · (2 − 1)
2 1 x
or
1 2
1 x
≤ dx ≤ 1
2 1 x
(Not a very good estimate)


Notes

Example
2
1
Evaluate dx.
1 x

Solution

2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1
= ln 2


12


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 over
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


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


13


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





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


14


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
ax
sin x dx = − cos x + C ax 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





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
3
Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1)
0
and (2, 3], and negative on (1, 2). If we want the area of the region, we
have to do
1 2 3
A= (x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx
0 1 2
1 3 1 2 3
= 3x − 2 x 2 + 2x
3
0
− 1 3
3 x − 3 x 2 + 2x
2 1
+ 1 3
3x − 2 x 2 + 2x
3
2
5 1 5 11
= − − + = .
6 6 6 6


15


Graph
Notes
y

x
1 2 3


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
3
Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1)
0
and (2, 3], and negative on (1, 2). If we want the area of the region, we
have to do
1 2 3
A= (x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx
0 1 2
1 3 1 2 3
= 3x − 2 x 2 + 2x
3
0
− 1 3
3x − 3 x 2 + 2x
2 1
+ 1 3
3x − 2 x 2 + 2x
3
2
5 1 5 11
= − − + = .
6 6 6 6


Interpretation of “negative area” in motion
Notes

There is an analog in rectlinear motion:
t1
v (t) dt is net distance traveled.
t0
t1
|v (t)| dt is total distance traveled.
t0


16


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

Second FTC:
b b
f (x) dx = F (x)
a a

where F is an antiderivative
of f .
Computes any “net change”
over an interval
Proving the FTC requires
the MVT


Notes

17

Lesson 24: Evaluating Definite Integrals (handout)

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