Lesson 24: Evaluating Definite Integrals (slides)

Section 5.3
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

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

Most got problems 1 and 3 right.

. . . . . .


Homework: The Bad (steel pipe)
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?

. . . . . .


Problem
around the corner?

6
.
.

. 9
.
. . . . . .


Problem
around the corner?

6
.

.
θ
.

. 9
.
. . . . . .


Problem
around the corner?
.
.
θ
6
.

.
θ
.

. 9
.
. . . . . .


Problem
around the corner?
.
.
θ
6
. 6
.

9
.

.
θ
.

. 9
.
. . . . . .


Problem
around the corner?
.
.
θ
6
. 6
.

9
.

.
θ cθ
. . cs
9

. 9
.
. . . . . .


Problem
around the corner?
.
.
θ
6
. 6
.
cθ
. se
6
9
.

.
θ cθ
. . cs
9

. 9
.
. . . . . .


Solution
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
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 ( a )2/3
b
min = a 1+ +b 1+
a b . . . . . .


Simplifying

√ √
( )2/3 ( a )2/3
b
min = a 1 + +b 1+
a b
√ √
b2/3 a2/3 a2/3 b2/3
=b + 2/3 + a +
b2/3 b a2/3 a2/3
√ √
b 2/3 a
= 1/3 b + a 2/3 +
1/3
a2/3 + b2/3
b √ a
√
2/3 2/3 2/3 + a2/3 a2/3 + b2/3
=b b +a
√
= (b + a ) b2/3 + a2/3
2/3 2/3

= (a2/3 + b2/3 )3/2

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


Homework: The Bad (Diving Board)

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) + 1 ρg(L − x)2
2

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.

. . . . . .


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

Antidifferentiating once gives

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

. . . . . .


Don't stop there!
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) = − 2 mg(L − x)2 − 1 ρg(L − x)3 + C
1
6

gives

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


Solution completed
So

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

which means
( )
1 1 1 1
EIy(L) = mgL2 + L3 L − mgL3 − ρgL4
2 6 6 24
1 1 1 1
= mgL3 + L4 − mgL3 − ρgL4
2 6 6 24
1 1
= mgL3 + ρgL4
3 8
3 ( )
gL m ρL
=⇒ y(L) = +
EI 3 8
. . . . . .


Homework: The Ugly

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.

. . . . . .


Objectives

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

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

. . . . . .



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

∫ 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

. . . . . .



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

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

. . . . . .


Definite Integrals We Know So Far

If the integral computes an
area and we know the
area, we can use that. For
instance,
y
.
∫ 1√
π
1 − x2 dx =
0 4

By brute force we
.
computed x
.
∫ 1 ∫ 1
1 1
x2 dx = x3 dx =
0 3 0 4

. . . . . .



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

. . . . . .


Example
∫ 2
1
Estimate dx using Property ??.
1 x

. . . . . .


Example
∫ 2
1
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 .
∫ 2 x
.
1 1
≤ dx ≤ 1
2 1 x
(Not a very good estimate)
. . . . . .


Outline


Examples




. . . . . .


Socratic dialogue

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 . . .

. . . . . .


Theorem of the Day

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

. . . . . .


Theorem of the Day

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

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 )

. . . . . .



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!

. . . . . .



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)

. . . . . .



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

Example

Find the area between y = x3
the x-axis, x = 0 and x = 1.

.

. . . . . .



Example


.

Solution

∫ 1 1
x4 1
A= x3 dx = =
0 4 0 4

. . . . . .



Example


.

Solution

∫ 1 1
x4 1
A= x3 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
Example
Find the area enclosed by the parabola y = x2 and y = 1.

. . . . . .


Example

.
1
.

. . .
−
. 1 1
.

. . . . . .


Example

.
1
.

. . .
−
. 1 1
.

Solution

∫ 1 [ ]1 ( [ )]
x3 1 1 4
A=2− x dx = 2 − 2
=2− − − =
−1 3 −1 3 3 3
. . . . . .


Computing exactly what we earlier estimated

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

. . . . . .


Example
∫ 1
4
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
. . . . . .



Example
∫ 1
4
0 1 + x2

Solution

∫ 1 ∫ 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2

. . . . . .



Example
∫ 1
4
0 1 + x2

Solution

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

. . . . . .



Example
∫ 1
4
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)

. . . . . .



Example
∫ 1
4
0 1 + x2

Solution

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

. . . . . .



Example
∫ 1
4
0 1 + x2

Solution

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

. . . . . .



Example
∫ 2
1
Evaluate dx.
1 x

. . . . . .


Example
∫ 2
1
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 .
∫ 2 x
.
1 1
≤ dx ≤ 1
2 1 x
(Not a very good estimate)
. . . . . .



Example
∫ 2
1
Evaluate dx.
1 x

Solution

∫ 2
1
dx
1 x

. . . . . .



Example
∫ 2
1
Evaluate dx.
1 x

Solution

∫ 2
1
dx = ln x|2
1
1 x

. . . . . .



Example
∫ 2
1
Evaluate dx.
1 x

Solution

∫ 2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1

. . . . . .



Example
∫ 2
1
Evaluate dx.
1 x

Solution

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

. . . . . .


Outline


Examples




. . . . . .



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:

. . . . . .



∫ b
F′ (x) dx = F(b) − F(a),
a


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

. . . . . .



∫ 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

. . . . . .



∫ 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


Examples




. . . . . .


A new notation for antiderivatives

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).

. . . . . .


A new notation for antiderivatives

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
∫
x2 dx = 1 x3 + C.
3

. . . . . .


.
∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx
∫ ∫ ∫
xn+1
xn dx = + C (n ̸= −1) cf(x) dx = c f(x) dx
n+1 ∫
∫
1
ex dx = ex + 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


Examples




. . . . . .


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

. . . . . .


Example

Solution
∫ 3
Consider (x − 1)(x − 2) dx.
0

. . . . . .


Graph

y
.

. . . . x
.
1
. 2
. 3
.

. . . . . .


Example

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).

. . . . . .


Example

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 [ ]2 [ ]3
= − 1 3
3x
3 2
2x + 2x − 1 x3 − 3 x2 + 2x + 1 x3 − 3 x2 + 2x
3 2 3 2
( ) 0 1 2
5 1 5 11
= − − + = .
6 6 6 6

. . . . . .


Interpretation of “negative area" in motion

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

. . . . . .


What about the constant?

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

But notice
[ 4 ]1 ( )
x 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

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

. . . . . .


Lesson 24: Evaluating Definite Integrals (slides)

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Lesson 24: Evaluating Definite Integrals (slides)