Lesson 24: Evaluating Definite Integrals (handout)

Section 5.3Evaluating Definite Integrals

V63.0121.006/016, Calculus I

New York University

April 20, 2010

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I April 30: Quiz 5 on §§5.1–5.4

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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 10:00am as previouslyannounced) 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.


Notes

Notes

Notes

1

Section 5.3 : Evaluating Definite IntegralsV63.0121.006/016, Calculus I April 20, 2010

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

Homework: The Bad (steel pipe)

Problem

A 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 isthe length of the longest pipe that can be carried horizontally around thecorner?

9

6

θ

θ

9

6

9 csc θ

6 sec θ


Solution

Solution

The 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 θ

cos2 θ− b

cos θ

sin2 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 =b

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

where tan3 θmin =b

a=⇒ tan θmin =

(b

a

)1/3

.

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

We get the minimum value is

min = a

√1 +

(b

a

)2/3

+ b

√1 +

(a

b

)2/3


Notes

Notes

Notes

2


Simplifying

min = a

√1 +

(b

a

)2/3

+ b

√1 +

(a

b

)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 +

a

a1/3

√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)

Problem

If a diver of mass m stands at the end of a diving board with length L andlinear density ρ, then the board takes on the shape of a curve y = 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 ofthe board.


Solution

We haveEIy ′′(x) = mg(L− x) + 1

2ρg(L− x)2

Antidifferentiating once gives

EIy ′(x) = −12 mg(L− x)2 − 1

6ρg(L− x)3 + C

Once more:

EIy(x) = 16 mg(L− x)3 + 1

24ρg(L− x)4 + Cx + D

where C and D are constants.


Notes

Notes

Notes

3


Don’t stop there!

Plugging y(0) = 0 into

EIy ′(x) = 16 mg(L− x)3 + 1

24ρg(L− x)4 + Cx + D

gives

0 = 16 mgL3 + 1

24ρgL4 + D =⇒ D = −1

6mgL3 − 1

24ρgL4

Plugging y ′(0) = 0 into

EIy ′(x) = −12 mg(L− x)2 − 1

6ρg(L− x)3 + C

gives

0 = −12 mgL2 − 1

6ρgL3 + C =⇒ C =1

2mgL2 +

1

6ρgL3


Solution completed

So

EIy(x) = 16 mg(L− x)3 + 1

24ρg(L− x)4

+

(1

2mgL2 +

1

6L3

)x − 1

6mgL3 − 1

24ρgL4

which means

EIy(L) =

(1

2mgL2 +

1

6L3

)L− 1

6mgL3 − 1

24ρgL4

=1

2mgL3 +

1

6L4 − 1

6mgL3 − 1

24ρgL4

=1

3mgL3 +

1

8ρgL4

=⇒ y(L) =gL3

EI

(m

3+ρL

8

)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 and arecopying answers word for word.

I This is very easy to catch: the graders are following the same solutionmanual.

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.


Notes

Notes

Notes

4


Objectives

I Use the Evaluation Theoremto evaluate definite integrals.

I Write antiderivatives asindefinite integrals.

I Interpret definite integrals as“net change” of a functionover 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

Definition

If 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 − a

n, and for each i , xi = a + i∆x , and ci is a point in

[xi−1, xi ].

Theorem

If 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 .


Notes

Notes

Notes

5


Notation/Terminology

∫ b

af (x) dx = lim

n→∞

n∑i=1

f (ci ) ∆x

I

∫— integral sign (swoopy S)

I f (x) — integrand

I 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.


Notes

Notes

Notes

6


Definite Integrals We Know So Far

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

0

√1− x2 dx =

π

4

I By brute force we computed∫ 1

0x2 dx =

1

3

∫ 1

0x3 dx =

1

4

x

y


Example

Estimate

∫ 1

0

4

1 + x2dx using the midpoint rule and four divisions.

Solution

Dividing up [0, 1] into 4 pieces gives

x0 = 0, x1 =1

4, x2 =

2

4, x3 =

3

4, x4 =

4

4

So the midpoint rule gives

M4 =1

4

(4

1 + (1/8)2+

4

1 + (3/8)2+

4

1 + (5/8)2+

4

1 + (7/8)2

)=

1

4

(4

65/64+

4

73/64+

4

89/64+

4

113/64

)=

150, 166, 784

47, 720, 465≈ 3.1468


Comparison Properties of the Integral

Theorem

Let 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)


Notes

Notes

Notes

7


Example

Estimate

∫ 2

1

1

xdx using Property 8.

SolutionSince

1 ≤ x ≤ 2 =⇒ 1

2≤ 1

x≤ 1

1

we have

1

2·(2−1) ≤

∫ 2

1

1

xdx ≤ 1 ·(2−1)

or1

2≤∫ 2

1

1

xdx ≤ 1

(Not a very good estimate)

x

y


Outline







Socratic dialogue

I The definite integral ofvelocity measuresdisplacement (net distance)

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 . . .


Notes

Notes

Notes

8


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).

Note

In Section 5.3, this theorem is called “The Evaluation Theorem”. Nobodyelse in the world calls it that.


Proving the Second FTC

Proof.

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

nas usual. For

each 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 if (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)


Notes

Notes

Notes

9



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

Solution

A =

∫ 1

0x3 dx =

x4

4

∣∣∣∣10

=1

4

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


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−[

1

3−(−1

3

)]=

4

3


Notes

Notes

Notes

10


Computing exactly what we earlier estimated

Example

Evaluate the integral

∫ 1

0

4

1 + x2dx .

Solution

∫ 1

0

4

1 + x2dx = 4

∫ 1

0

1

1 + x2dx

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

= 4(π

4− 0)

= π


Example

Estimate

∫ 1

0

4

1 + x2dx using the midpoint rule and four divisions.

Solution

Dividing up [0, 1] into 4 pieces gives

x0 = 0, x1 =1

4, x2 =

2

4, x3 =

3

4, x4 =

4

4

So the midpoint rule gives

M4 =1

4

(4

1 + (1/8)2+

4

1 + (3/8)2+

4

1 + (5/8)2+

4

1 + (7/8)2

)=

1

4

(4

65/64+

4

73/64+

4

89/64+

4

113/64

)=

150, 166, 784

47, 720, 465≈ 3.1468



Example

Evaluate the integral

∫ 1

0

4

1 + x2dx .

Solution

∫ 1

0

4

1 + x2dx = 4

∫ 1

0

1

1 + x2dx

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

= 4(π

4− 0)

= π


Notes

Notes

Notes

11



Example

Evaluate

∫ 2

1

1

xdx .

Solution

∫ 2

1

1

xdx = ln x |21

= ln 2− ln 1

= ln 2


Example

Estimate

∫ 2

1

1

xdx using Property 8.

SolutionSince

1 ≤ x ≤ 2 =⇒ 1

2≤ 1

x≤ 1

1

we have

1

2·(2−1) ≤

∫ 2

1

1

xdx ≤ 1 ·(2−1)

or1

2≤∫ 2

1

1

xdx ≤ 1

(Not a very good estimate)

x

y



Example

Evaluate

∫ 2

1

1

xdx .

Solution

∫ 2

1

1

xdx = ln x |21

= ln 2− ln 1

= ln 2


Notes

Notes

Notes

12


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:

Theorem

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

t0

v(t) dt = s(t1)− s(t0).




aF ′(x) dx = F (b)− F (a),


Theorem

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

C (x) = C (0) +

∫ x

0MC (q) dq.


Notes

Notes

Notes

13




aF ′(x) dx = F (b)− F (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

m(x) =

∫ x

0ρ(s) ds.


Outline







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

3 x3 + C .


Notes

Notes

Notes

14


My first table of integrals

∫[f (x) + g(x)] dx =

∫f (x) dx +

∫g(x) dx∫

xn dx =xn+1

n + 1+ C (n 6= −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∫

1

xdx = 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, and thevertical 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

=[

13 x3 − 3

2 x2 + 2x]1

0−[

13 x3 − 3

2 x2 + 2x]2

1+[

13 x3 − 3

2 x2 + 2x]3

2

=5

6−(−1

6

)+

5

6=

11

6.


Notes

Notes

Notes

15


Graph

x

y

1 2 3


Example

Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and thevertical 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

=[

13 x3 − 3

2 x2 + 2x]1

0−[

13 x3 − 3

2 x2 + 2x]2

1+[

13 x3 − 3

2 x2 + 2x]3

2

=5

6−(−1

6

)+

5

6=

11

6.


Interpretation of “negative area” in motion

There is an analog in rectlinear motion:

I

∫ t1

t0

v(t) dt is net distance traveled.

I

∫ t1

t0

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


Notes

Notes

Notes

16


What about the constant?

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

0x3 dx =

x4

4

∣∣∣∣10

=1

4− 0 =

1

4

I But notice[x4

4+ C

]1

0

=

(1

4+ C

)− (0 + C ) =

1

4+ C − C =

1

4

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 an antiderivativeof f .

I Computes any “net change”over an interval

I Proving the FTC requiresthe MVT


Notes

Notes

Notes

17


Education

Lesson 24: Evaluating Definite Integrals (handout)