Lesson 25: Evaluating Definite Integrals (Section 041 slides)

..

Section 5.3Evaluating Definite Integrals

V63.0121.041, Calculus I

New York University

December 6, 2010

Announcements

I Today: Section 5.3I Wednesday: Section 5.4I Monday, December 13: Section 5.5I ”Monday,” December 15: Review and Movie Day!I Monday, December 20, 12:00–1:50pm: Final Exam

. . . . . .

. . . . . .

Announcements

I Today: Section 5.3I Wednesday: Section 5.4I Monday, December 13:Section 5.5

I ”Monday,” December 15:Review and Movie Day!

I Monday, December 20,12:00–1:50pm: Final Exam

V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 2 / 41

. . . . . .

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 IntegralsExamples

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

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


. . . . . .

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

Estimate∫ 1

0

41+ x2



x0 = 0, x1 =14, x2 =

24, x3 =

34, x4 =

44


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

Estimate∫ 1

0

41+ x2



x0 = 0, x1 =14, x2 =

24, x3 =

34, x4 =

44


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

Estimate∫ 1

0

41+ x2



x0 = 0, x1 =14, x2 =

24, x3 =

34, x4 =

44


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


. . . . . .

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

Theorem

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c

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


. . . . . .

Illustrating Property 5

Theorem

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c


..x

.

y

..a

..b

..c


. . . . . .


Theorem

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c


..x

.

y

..a

..b

..c

.

∫ b

af(x)dx


. . . . . .


Theorem

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c


..x

.

y

..a

..b

..c

.

∫ b

af(x)dx

.

∫ c

bf(x)dx


. . . . . .


Theorem

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c


..x

.

y

..a

..b

..c

.

∫ b

af(x)dx

.

∫ c

bf(x)dx

.

∫ c

af(x)dx


. . . . . .


Theorem

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c


..x

.

y

..a..

b..

c


. . . . . .


Theorem

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c


..x

.

y

..a..

b..

c.

∫ b

af(x)dx


. . . . . .


Theorem

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c


..x

.

y

..a..

b..

c.

∫ c

bf(x)dx =

−∫ b

cf(x)dx


. . . . . .


Theorem

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c


..x

.

y

..a..

b..

c.

∫ c

bf(x)dx =

−∫ b

cf(x)dx

.

∫ c

af(x)dx


. . . . . .

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)


. . . . . .




af(x) dx ≥ 0


af(x)dx ≥

∫ b

ag(x)dx


m(b− a) ≤∫ b



. . . . . .




af(x) dx ≥ 0


af(x)dx ≥

∫ b

ag(x)dx


m(b− a) ≤∫ b



. . . . . .

Estimating an integral with inequalities

Example

Estimate∫ 2

1

1xdx using Property 8.

SolutionSince

1 ≤ x ≤ 2 =⇒ 12≤ 1

x≤ 1

1we have

12· (2− 1) ≤

∫ 2

1

1xdx ≤ 1 · (2− 1)

or12≤

∫ 2

1

1xdx ≤ 1


. . . . . .


Example

Estimate∫ 2

1


SolutionSince

1 ≤ x ≤ 2 =⇒ 12≤ 1

x≤ 1

1we have

12· (2− 1) ≤

∫ 2

1

1xdx ≤ 1 · (2− 1)

or12≤

∫ 2

1

1xdx ≤ 1


. . . . . .

Outline







. . . . . .

Socratic proof

I The definite integral ofvelocity measuresdisplacement (netdistance)

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).

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

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

as usual.

I For 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)

xi − xi−1= F′(ci) = f(ci)

Orf(ci)∆x = F(xi)− F(xi−1)


. . . . . .



as usual.


F(xi)− F(xi−1)




. . . . . .



as usual.


F(xi)− F(xi−1)




. . . . . .

Proof continued

I We have for each i

f(ci)∆x = F(xi)− F(xi−1)

I 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)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof continued




Sn =n∑

i=1

f(ci)∆x =n∑

i=1

(F(xi)− F(xi−1))


= F(xn)− F(x0) = F(b)− F(a)


. . . . . .

Proof Completed

I We have shown for each n,

Sn = F(b)− F(a)

Which does not depend on n.

I So in the limit∫ b

af(x)dx = lim

n→∞Sn = lim

n→∞(F(b)− F(a)) = F(b)− F(a)


. . . . . .

Proof Completed

I We have shown for each n,

Sn = F(b)− F(a)

Which does not depend on n.I 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=

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 .



. . . . . .


Example

Find the area enclosed by the parabola y = x2 and the line 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 an integral we estimated before

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



x0 = 0, x1 =14, x2 =

24, x3 =

34, x4 =

44


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

1we have

12· (2− 1) ≤

∫ 2

1

1xdx ≤ 1 · (2− 1)

or12≤

∫ 2

1

1xdx ≤ 1


. . . . . .


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 changebetween the sides of that 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 of the region bounded by the lines x = 1, x = 4, thex-axis, and the curve y = ex.

SolutionThe answer is ∫ 4

1ex dx = ex|41 = e4 − e.


. . . . . .


Example

Find the area of the region bounded by the lines x = 1, x = 4, thex-axis, and the curve y = ex.

SolutionThe answer is ∫ 4

1ex dx = ex|41 = e4 − e.


. . . . . .


Example

Find the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.

Solution

I The answer is∫ 1

0arcsin x dx, but we

do not know an antiderivative forarcsin.

I Instead compute the area as

π

2−∫ π/2

0sin y dy =

π

2−[− cos x]π/20 =

π

2−1

..x

.

y

..1

..

π/2


. . . . . .


Example


Solution





π

2−∫ π/2

0sin y dy =

π

2−[− cos x]π/20 =

π

2−1

..x

.

y

..1

..

π/2


. . . . . .


Example


Solution





π

2−∫ π/2

0sin y dy

=π

2−[− cos x]π/20 =

π

2−1

..x

.

y

..1

..

π/2


. . . . . .


Example


Solution





π

2−∫ π/2

0sin y dy =

π

2−[− cos x]π/20

=π

2−1

..x

.

y

..1

..

π/2


. . . . . .


Example


Solution





π

2−∫ π/2

0sin y dy =

π

2−[− cos x]π/20 =

π

2−1 ..

x.

y

..1

..

π/2


. . . . . .

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 is

immaterial.


. . . . . .

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 25: Evaluating Definite Integrals (Section 041 slides)