Lesson 26: Evaluating Definite Integrals

Section 5.3Evaluating Definite Integrals

V63.0121.002.2010Su, Calculus I

New York University

June 21, 2010

Announcements

I Final Exam is Thursday in class

Announcements

I Sections 5.3–5.4 today

I Section 5.5 tomorrow

I Review and Movie DayWednesday

I Final exam ThursdayI roughly half-and-half

MC/FRI FR is all post-midtermI MC might have some

pre-midterm stuff on it

V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 2 / 44

Resurrection Policy

If your final score beats your midterm score, we will add 10% to its weight,and subtract 10% from the midterm weight.

Image credit: Scott Beale / Laughing SquidV63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 3 / 44

http://laughingsquid.com/

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 IntegralsExamples

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 .


Notation/Terminology

∫ b

af (x) dx

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


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

Estimate

∫ 1

0

4


Solution


x0 = 0, x1 =1

4, x2 =

2

4, x3 =

3

4, x4 =

4

4


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

Estimate

∫ 1

0

4


Solution


x0 = 0, x1 =1

4, x2 =

2

4, x3 =

3

4, x4 =

4

4


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

Estimate

∫ 1

0

4


Solution


x0 = 0, x1 =1

4, x2 =

2

4, x3 =

3

4, x4 =

4

4


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


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 the area,we can use that. Forinstance,∫ 1

0

√1− x2 dx =

π

2

I By brute force we computed∫ 1

0x2 dx =

1

3

∫ 1

0x3 dx =

1

4

x

y


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)


Integral of a nonnegative function is nonnegative

Proof.

If f (x) ≥ 0 for all x in [a, b], then for any number of divisions n and choiceof sample points {ci}:

Sn =n∑

i=1

f (ci )︸︷︷︸≥0

∆x ≥n∑

i=1

0 ·∆x = 0

Since Sn ≥ 0 for all n, the limit of {Sn} is nonnegative, too:∫ b

af (x) dx = lim

n→∞Sn︸︷︷︸≥0

≥ 0



Theorem



af (x) dx ≥ 0


af (x) dx ≥

∫ b

ag(x) dx


m(b − a) ≤∫ b



The definite integral is “increasing”

Proof.

Let h(x) = f (x)− g(x). If f (x) ≥ g(x) for all x in [a, b], then h(x) ≥ 0for all x in [a, b]. So by the previous property∫ b

ah(x) dx ≥ 0

This means that∫ b

af (x) dx −

∫ b

ag(x) dx =

∫ b

a(f (x)− g(x)) dx =

∫ b

ah(x) dx ≥ 0

So ∫ b

af (x) dx ≥

∫ b

ag(x) dx



Theorem



af (x) dx ≥ 0


af (x) dx ≥

∫ b

ag(x) dx


m(b − a) ≤∫ b



Bounding the integral using bounds of the function

Proof.

If m ≤ f (x) ≤ M on for all x in [a, b], then by the previous property∫ b

am dx ≤

∫ b

af (x) dx ≤

∫ b

aM dx

By Property 1, the integral of a constant function is the product of theconstant and the width of the interval. So:

m(b − a) ≤∫ b



Estimating an integral with inequalities

Example

Estimate

∫ 2

1

1

xdx using Property 8.

Solution

Since

1 ≤ x ≤ 2 =⇒ 1

2≤ 1

x≤ 1

1

we have1

2· (2− 1) ≤

∫ 2

1

1

xdx ≤ 1 · (2− 1)

or1

2≤∫ 2

1

1

xdx ≤ 1



Example

Estimate

∫ 2

1

1


Solution

Since

1 ≤ x ≤ 2 =⇒ 1

2≤ 1

x≤ 1

1

we have1

2· (2− 1) ≤

∫ 2

1

1

xdx ≤ 1 · (2− 1)

or1

2≤∫ 2

1

1

xdx ≤ 1


Outline







Socratic proof

I The definite integral ofvelocity measuresdisplacement (net distance)

I The derivative ofdisplacement is velocity

I So we can computedisplacement with thedefinite integral or theantiderivative 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).

Note

In Section 5.3, this theorem is called “The Evaluation Theorem”. Nobodyelse 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).

Note

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


Proving the Second FTC

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)


Proof continued

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)

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


Proof continued

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)

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


Proof Completed

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)


Computing area with the Second FTC

Example

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



Example


Solution

A =

∫ 1

0x3 dx =

x4

4

∣∣∣∣10

=1

4




Example


Solution

A =

∫ 1

0x3 dx =

x4

4

∣∣∣∣10

=1

4




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



Example


−1 1

1

Solution

A = 2−∫ 1

−1x2 dx = 2−

[x3

3

]1

−1

= 2−[

1

3−(−1

3

)]=

4

3



Example


−1 1

1

Solution

A = 2−∫ 1

−1x2 dx = 2−

[x3

3

]1

−1

= 2−[

1

3−(−1

3

)]=

4

3


Computing an integral we estimated before

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


Solution


x0 = 0, x1 =1

4, x2 =

2

4, x3 =

3

4, x4 =

4

4


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


∫ 1

0

4

1 + x2dx .

Solution

∫ 1

0

4

1 + x2dx = 4

∫ 1

0

1

1 + x2dx


= 4(π

4− 0)

= π



Example


∫ 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


∫ 1

0

4

1 + x2dx .

Solution

∫ 1

0

4

1 + x2dx = 4

∫ 1

0

1

1 + x2dx


= 4(π

4− 0)

= π



Example


∫ 1

0

4

1 + x2dx .

Solution

∫ 1

0

4

1 + x2dx = 4

∫ 1

0

1

1 + x2dx


= 4(π

4− 0)

= π



Example


∫ 1

0

4

1 + x2dx .

Solution

∫ 1

0

4

1 + x2dx = 4

∫ 1

0

1

1 + x2dx


= 4(π

4− 0)

= π



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


Solution

Since

1 ≤ x ≤ 2 =⇒ 1

2≤ 1

x≤ 1

1

we have1

2· (2− 1) ≤

∫ 2

1

1

xdx ≤ 1 · (2− 1)

or1

2≤∫ 2

1

1

xdx ≤ 1



Example

Evaluate

∫ 2

1

1

xdx .

Solution

∫ 2

1

1

xdx

= ln x |21

= ln 2− ln 1

= ln 2



Example

Evaluate

∫ 2

1

1

xdx .

Solution

∫ 2

1

1

xdx = ln x |21

= ln 2− ln 1

= ln 2



Example

Evaluate

∫ 2

1

1

xdx .

Solution

∫ 2

1

1

xdx = ln x |21

= ln 2− ln 1

= ln 2



Example

Evaluate

∫ 2

1

1

xdx .

Solution

∫ 2

1

1

xdx = 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 betweenthe sides of that interval. This has many ramifications:




aF ′(x) dx = F (b)− F (a),


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.




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 .


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 .


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.


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

=[

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.


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

=[

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.


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

=[

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.


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 The second Fundamental Theorem of Calculus:∫ b

af (x) dx = F (b)− F (a)

where F ′ = f .

I Definite integrals represent net change of a function over an interval.

I We write antiderivatives as indefinite integrals

∫f (x) dx


Technology

Lesson 26: Evaluating Definite Integrals