Lesson 24: Evaluating Definite Integrals (slides)

.

.

Section 5.3Evaluating Definite Integrals

V63.0121.006/016, Calculus I

New York University

April 20, 2010

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

..Image credit: docman . . . . . .

http://flickr.com/photos/docman/

. . . . . .

Announcements

I April 16: Quiz 4 on§§4.1–4.4

I April 29: Movie Day!!I April 30: Quiz 5 on§§5.1–5.4

I Monday, May 10,12:00noon (not 10:00amas 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)

ProblemA steel pipe is being carried down a hallway 9 ft wide. At the end of thehall 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 horizontallyaround the corner?

. .9

.6

.

..θ

..θ

.9

.6

.9 cscθ

.6 secθ


. . . . . .



. .9

.6

.

..θ

..θ

.9

.6

.9 cscθ

.6 secθ


. . . . . .



. .9

.6

.

..θ

..θ

.9

.6

.9 cscθ

.6 secθ


. . . . . .



. .9

.6.

..θ

..θ

.9

.6

.9 cscθ

.6 secθ


. . . . . .



. .9

.6

.

..θ

..θ

.9

.6

.9 cscθ

.6 secθ


. . . . . .



. .9

.6

.

..θ

..θ

.9

.6

.9 cscθ

.6 secθ


. . . . . .



. .9

.6

.

..θ

..θ

.9

.6

.9 cscθ

.6 secθ


. . . . . .



. .9

.6

.

..θ

..θ

.9

.6

.9 cscθ

.6 secθ


. . . . . .



. .9

.6

.

..θ

..θ

.9

.6

.9 cscθ

.6 secθ


. . . . . .

Solution

SolutionThe longest pipe that barely fits is the smallest pipe that almost doesn’tfit. 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

= asin θcos2 θ

− bcos θsin2 θ

=a sin3 θ − b cos3 θ

sin2 θ cos2 θ

So the critical point is when

a sin3 θ = b cos3 θ =⇒ tan3 θ =ba


. . . . . .

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

where tan3 θmin =ba

=⇒ tan θmin =

(ba

)1/3.

Using1+ tan2 θ = sec2 θ 1+ cot2 θ = csc2 θ

We get the minimum value is

min = a

√1+

(ba

)2/3+ b

√1+

(ab

)2/3V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 6 / 48

. . . . . .

Simplifying

min = a

√1+

(ba

)2/3+ b

√1+

(ab

)2/3= b

√b2/3

b2/3+

a2/3

b2/3+ a

√a2/3

a2/3+

b2/3

a2/3

=b

b1/3

√b2/3 + a2/3 +

aa1/3

√a2/3 + b2/3

= b2/3√

b2/3 + a2/3 + a2/3√

a2/3 + b2/3

= (b2/3 + a2/3)√

b2/3 + a2/3

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

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


. . . . . .

Homework: The Bad (Diving Board)

ProblemIf a diver of mass m stands at the end of a diving board with length Land linear density ρ, then the board takes on the shape of a curvey = f(x), where

EIy′′ = mg(L− x) + 12ρg(L− x)2

E and I are positive constants that depend of the material of the boardand 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.


. . . . . .

SolutionWe have

EIy′′(x) = mg(L− x) + 12ρg(L− x)2

Antidifferentiating once gives

EIy′(x) = −12mg(L− x)2 − 1

6ρg(L− x)3 + C

Once more:

EIy(x) = 16mg(L− x)3 + 1

24ρg(L− x)4 + Cx+ D

where C and D are constants.


. . . . . .

Don't stop there!

Plugging y(0) = 0 into

EIy′(x) = 16mg(L− x)3 + 1

24ρg(L− x)4 + Cx+ D

gives

0 = 16mgL3 + 1

24ρgL4 + D =⇒ D = −1

6mgL3 − 1

24ρgL4

Plugging y′(0) = 0 into

EIy′(x) = −12mg(L− x)2 − 1

6ρg(L− x)3 + C

gives

0 = −12mgL2 − 1

6ρgL3 + C =⇒ C =

12mgL2 +

16ρgL3


. . . . . .

Solution completed

So

EIy(x) = 16mg(L− x)3 + 1

24ρg(L− x)4

+

(12mgL2 +

16L3

)x− 1

6mgL3 − 1

24ρgL4

which means

EIy(L) =(12mgL2 +

16L3

)L− 1

6mgL3 − 1

24ρgL4

=12mgL3 +

16L4 − 1

6mgL3 − 1

24ρgL4

=13mgL3 +

18ρgL4

=⇒ y(L) =gL3

EI

(m3

+ρL8

)V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 11 / 48

. . . . . .

Homework: The Ugly

I Some students have gotten their hands on a solution manual andare copying answers word for word.

I This is very easy to catch: the graders are following the samesolution manual.

I This is not very productive: the best you will do is ace 10% of yourcourse grade.

I This is a violation of academic integrity. I do not take it lightly.


. . . . . .

Objectives

I Use the EvaluationTheorem to evaluatedefinite integrals.

I Write antiderivatives asindefinite integrals.

I Interpret definite integralsas “net change” of afunction over an interval.


. . . . . .

Outline

Last time: The Definite IntegralThe definite integral as a limitProperties of the integralComparison Properties of the Integral

Evaluating Definite IntegralsThe Theorem of the DayExamples

The Integral as Total Change

Indefinite IntegralsMy first table of integrals

Computing Area with integrals


. . . . . .

The definite integral as a limit

DefinitionIf f is a function defined on [a,b], the definite integral of f from a to bis the number ∫ b

af(x)dx = lim

n→∞

n∑i=1

f(ci)∆x

where ∆x =b− an

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

TheoremIf f is continuous on [a,b] or if f has only finitely many jumpdiscontinuities, then f is integrable on [a,b]; that is, the definite integral∫ b

af(x) dx exists and is the same for any choice of ci.


. . . . . .

Notation/Terminology

∫ b

af(x)dx = lim

n→∞

n∑i=1

f(ci)∆x

I∫

— integral sign (swoopy S)

I f(x) — integrandI a and b — limits of integration (a is the lower limit and b theupper limit)

I dx — ??? (a parenthesis? an infinitesimal? a variable?)I The process of computing an integral is called integration


. . . . . .

Properties of the integral

Theorem (Additive Properties of the Integral)

Let f and g be integrable functions on [a,b] and c a constant. Then

1.∫ b

ac dx = c(b− a)

2.∫ b

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

∫ b

af(x)dx+

∫ b

ag(x)dx.

3.∫ b

acf(x)dx = c

∫ b

af(x)dx.

4.∫ b

a[f(x)− g(x)] dx =

∫ b

af(x)dx−

∫ b

ag(x)dx.


. . . . . .

More Properties of the Integral

Conventions: ∫ a

bf(x)dx = −

∫ b

af(x)dx∫ a

af(x)dx = 0

This allows us to have

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c

bf(x)dx for all a, b, and c.


. . . . . .

Definite Integrals We Know So Far

I If the integral computes anarea and we know thearea, we can use that. Forinstance,∫ 1

0

√1− x2 dx =

π

4

I By brute force wecomputed∫ 1

0x2 dx =

13

∫ 1

0x3 dx =

14

..x

.y


. . . . . .

Comparison Properties of the Integral

TheoremLet f and g be integrable functions on [a,b].

6. If f(x) ≥ 0 for all x in [a,b], then∫ b

af(x) dx ≥ 0

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

af(x)dx ≥

∫ b

ag(x)dx

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

m(b− a) ≤∫ b

af(x)dx ≤ M(b− a)


. . . . . .

Example

Estimate∫ 2

1

1xdx using Property ??.

SolutionSince

1 ≤ x ≤ 2 =⇒ 12≤ 1

x≤ 1

1

we have

12·(2−1) ≤

∫ 2

1

1xdx ≤ 1·(2−1)

or12≤

∫ 2

1

1xdx ≤ 1

(Not a very good estimate)

..x

.y


. . . . . .

Example

Estimate∫ 2

1


SolutionSince

1 ≤ x ≤ 2 =⇒ 12≤ 1

x≤ 1

1

we have

12·(2−1) ≤

∫ 2

1

1xdx ≤ 1·(2−1)

or12≤

∫ 2

1

1xdx ≤ 1


..x

.y


. . . . . .

Outline







. . . . . .

Socratic dialogue

I The definite integral ofvelocity measuresdisplacement (netdistance)

I The derivative ofdisplacement is velocity

I So we can computedisplacement with thedefinite integral or anantiderivative of velocity

I But any function can be avelocity 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

af(x)dx = F(b)− F(a).

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


. . . . . .

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

af(x)dx = F(b)− F(a).

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


. . . . . .

Proving the Second FTC

Proof.

Divide up [a,b] into n pieces of equal width ∆x =b− an

as usual. Foreach i, F is continuous on [xi−1, xi] and differentiable on (xi−1, xi). Sothere is a point ci in (xi−1, xi) with

F(xi)− F(xi−1)

xi − xi−1= F′(ci) = f(ci)

Orf(ci)∆x = F(xi)− F(xi−1)

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


. . . . . .


Proof.

Divide up [a,b] into n pieces of equal width ∆x =b− an

as usual. Foreach i, F is continuous on [xi−1, xi] and differentiable on (xi−1, xi). Sothere is a point ci in (xi−1, xi) with

F(xi)− F(xi−1)

xi − xi−1= F′(ci) = f(ci)

Orf(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:

Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−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

af(x)dx = lim

n→∞Sn = lim

n→∞(F(b)− F(a)) = F(b)− F(a)


. . . . . .

Verifying earlier computations

Example

Find the area between y = x3

the x-axis, x = 0 and x = 1.

.

Solution

A =

∫ 1

0x3 dx =

x4

4

∣∣∣∣10=

14

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


. . . . . .


Example



.

Solution

A =

∫ 1

0x3 dx =

x4

4

∣∣∣∣10=

14



. . . . . .


Example



.

Solution

A =

∫ 1

0x3 dx =

x4

4

∣∣∣∣10=

14



. . . . . .

Verifying Archimedes

Example

Find the area enclosed by the parabola y = x2 and y = 1.

...−1

..1

..1

Solution

A = 2−∫ 1

−1x2 dx = 2−

[x3

3

]1−1

= 2−[13−

(−13

)]=

43


. . . . . .


Example


...−1

..1

..1

Solution

A = 2−∫ 1

−1x2 dx = 2−

[x3

3

]1−1

= 2−[13−

(−13

)]=

43


. . . . . .


Example


...−1

..1

..1

Solution

A = 2−∫ 1

−1x2 dx = 2−

[x3

3

]1−1

= 2−[13−

(−13

)]=

43


. . . . . .

Computing exactly what we earlier estimated

Example

Evaluate the integral∫ 1

0

41+ x2

dx.

Solution

∫ 1

0

41+ x2

dx = 4∫ 1

0

11+ x2

dx

= 4 arctan(x)|10= 4 (arctan 1− arctan 0)

= 4(π4− 0

)

= π


. . . . . .

Example

Estimate∫ 1

0

41+ x2

dx using the midpoint rule and four divisions.

SolutionDividing up [0,1] into 4 pieces gives

x0 = 0, x1 =14, x2 =

24, x3 =

34, x4 =

44

So the midpoint rule gives

M4 =14

(4

1+ (1/8)2+

41+ (3/8)2

+4

1+ (5/8)2+

41+ (7/8)2

)=

14

(4

65/64+

473/64

+4

89/64+

4113/64

)=

150,166,78447,720,465

≈ 3.1468


. . . . . .


Example


0

41+ x2

dx.

Solution

∫ 1

0

41+ x2

dx = 4∫ 1

0

11+ x2

dx


= 4(π4− 0

)

= π


. . . . . .


Example


0

41+ x2

dx.

Solution

∫ 1

0

41+ x2

dx = 4∫ 1

0

11+ x2

dx

= 4 arctan(x)|10

= 4 (arctan 1− arctan 0)

= 4(π4− 0

)

= π


. . . . . .


Example


0

41+ x2

dx.

Solution

∫ 1

0

41+ x2

dx = 4∫ 1

0

11+ x2

dx


= 4(π4− 0

)

= π


. . . . . .


Example


0

41+ x2

dx.

Solution

∫ 1

0

41+ x2

dx = 4∫ 1

0

11+ x2

dx


= 4(π4− 0

)

= π


. . . . . .


Example


0

41+ x2

dx.

Solution

∫ 1

0

41+ x2

dx = 4∫ 1

0

11+ x2

dx


= 4(π4− 0

)= π


. . . . . .


Example

Evaluate∫ 2

1

1xdx.

Solution

∫ 2

1

1xdx

= ln x|21

= ln 2− ln 1= ln 2


. . . . . .

Example

Estimate∫ 2

1


SolutionSince

1 ≤ x ≤ 2 =⇒ 12≤ 1

x≤ 1

1

we have

12·(2−1) ≤

∫ 2

1

1xdx ≤ 1·(2−1)

or12≤

∫ 2

1

1xdx ≤ 1


..x

.y


. . . . . .


Example

Evaluate∫ 2

1

1xdx.

Solution

∫ 2

1

1xdx

= ln x|21

= ln 2− ln 1= ln 2


. . . . . .


Example

Evaluate∫ 2

1

1xdx.

Solution

∫ 2

1

1xdx = ln x|21

= ln 2− ln 1= ln 2


. . . . . .


Example

Evaluate∫ 2

1

1xdx.

Solution

∫ 2

1

1xdx = ln x|21

= ln 2− ln 1

= ln 2


. . . . . .


Example

Evaluate∫ 2

1

1xdx.

Solution

∫ 2

1

1xdx = ln x|21

= ln 2− ln 1= ln 2


. . . . . .

Outline







. . . . . .


Another way to state this theorem is:∫ b

aF′(x)dx = F(b)− F(a),

or, the integral of a derivative along an interval is the total change overthat interval. This has many ramifications:


. . . . . .





TheoremIf v(t) represents the velocity of a particle moving rectilinearly, then∫ t1

t0v(t)dt = s(t1)− s(t0).


. . . . . .





TheoremIf MC(x) represents the marginal cost of making x units of a product,then

C(x) = C(0) +∫ x

0MC(q)dq.


. . . . . .





TheoremIf ρ(x) represents the density of a thin rod at a distance of x from itsend, then the mass of the rod up to x is

m(x) =∫ x

0ρ(s)ds.


. . . . . .

Outline







. . . . . .

A new notation for antiderivatives

To emphasize the relationship between antidifferentiation andintegration, we use the indefinite integral notation∫

f(x)dx

for any function whose derivative is f(x).

Thus∫x2 dx = 1

3x3 + C.


. . . . . .

A new notation for antiderivatives

To emphasize the relationship between antidifferentiation andintegration, we use the indefinite integral notation∫

f(x)dx

for any function whose derivative is f(x). Thus∫x2 dx = 1

3x3 + C.


. . . . . .

My first table of integrals.

.

∫[f(x) + g(x)] dx =

∫f(x)dx+

∫g(x)dx∫

xn dx =xn+1

n+ 1+ C (n ̸= −1)∫

ex dx = ex + C∫sin x dx = − cos x+ C∫cos x dx = sin x+ C∫sec2 x dx = tan x+ C∫

sec x tan x dx = sec x+ C∫1

1+ x2dx = arctan x+ C

∫cf(x)dx = c

∫f(x)dx∫

1xdx = ln |x|+ C∫

ax dx =ax

ln a+ C∫

csc2 x dx = − cot x+ C∫csc x cot x dx = − csc x+ C∫

1√1− x2

dx = arcsin x+ C


. . . . . .

Outline







. . . . . .

Example

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

Solution

Consider∫ 3

0(x− 1)(x− 2) dx. Notice the integrand is positive on [0,1)

and (2,3], and negative on (1,2). If we want the area of the region, wehave to do

A =

∫ 1

0(x− 1)(x− 2)dx−

∫ 2

1(x− 1)(x− 2)dx+

∫ 3

2(x− 1)(x− 2)dx

=[13x

3 − 32x

2 + 2x]10−

[13x

3 − 32x

2 + 2x]21+

[13x

3 − 32x

2 + 2x]32

=56−

(−16

)+

56=

116.


. . . . . .

Example


Solution

Consider∫ 3

0(x− 1)(x− 2) dx.

Notice the integrand is positive on [0,1)


A =

∫ 1

0(x− 1)(x− 2)dx−

∫ 2

1(x− 1)(x− 2)dx+

∫ 3

2(x− 1)(x− 2)dx

=[13x

3 − 32x

2 + 2x]10−

[13x

3 − 32x

2 + 2x]21+

[13x

3 − 32x

2 + 2x]32

=56−

(−16

)+

56=

116.


. . . . . .

Graph

. .x

.y

..1

..2

..3


. . . . . .

Example


Solution

Consider∫ 3


and (2,3], and negative on (1,2).

If we want the area of the region, wehave to do

A =

∫ 1

0(x− 1)(x− 2)dx−

∫ 2

1(x− 1)(x− 2)dx+

∫ 3

2(x− 1)(x− 2)dx

=[13x

3 − 32x

2 + 2x]10−

[13x

3 − 32x

2 + 2x]21+

[13x

3 − 32x

2 + 2x]32

=56−

(−16

)+

56=

116.


. . . . . .

Example


Solution

Consider∫ 3



A =

∫ 1

0(x− 1)(x− 2)dx−

∫ 2

1(x− 1)(x− 2)dx+

∫ 3

2(x− 1)(x− 2)dx

=[13x

3 − 32x

2 + 2x]10−

[13x

3 − 32x

2 + 2x]21+

[13x

3 − 32x

2 + 2x]32

=56−

(−16

)+

56=

116.


. . . . . .

Interpretation of “negative area" in motion

There is an analog in rectlinear motion:

I∫ t1

t0v(t)dt is net distance traveled.

I∫ t1

t0|v(t)|dt is total distance traveled.


. . . . . .

What about the constant?

I It seems we forgot about the +C when we say for instance∫ 1

0x3 dx =

x4

4

∣∣∣∣10=

14− 0 =

14

I But notice[x4

4+ C

]10=

(14+ C

)− (0+ C) =

14+ C− C =

14

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


. . . . . .

Summary

I Second FTC:∫ b

af(x)dx = F(x)

∣∣∣∣ba

where F is anantiderivative of f.

I Computes any “netchange” over an interval

I Proving the FTC requiresthe MVT


Documents

Lesson 24: Evaluating Definite Integrals (slides)