Lesson 27: The Fundamental Theorem of Calculus

Section 5.4The Fundamental Theorem of Calculus

V63.0121.002.2010Su, Calculus I

New York University

June 21, 2010

Announcements

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Announcements

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V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 2 / 33

Objectives

I State and explain theFundemental Theorems ofCalculus

I Use the first fundamentaltheorem of calculus to findderivatives of functionsdefined as integrals.

I Compute the average valueof an integrable functionover a closed interval.


Outline

Recall: The Evaluation Theorem a/k/a 2FTC

The First Fundamental Theorem of CalculusThe Area FunctionStatement and proof of 1FTCBiographies

Differentiation of functions defined by integrals“Contrived” examplesErfOther applications


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

∆x→0

n∑i=1

f (ci ) ∆x


Big time Theorem

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).


The Integral as Total Change

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.


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





An area function

Let f (t) = t3 and define g(x) =

∫ x

0f (t) dt. Can we evaluate the integral

in g(x)?

0 x

Dividing the interval [0, x ] into n pieces

gives ∆t =x

nand ti = 0 + i∆t =

ix

n. So

Rn =x

n· x3

n3+

x

n· (2x)3

n3+ · · ·+ x

n· (nx)3

n3

=x4

n4

(13 + 23 + 33 + · · ·+ n3

)=

x4

n4

[12 n(n + 1)

]2=

x4n2(n + 1)2

4n4→ x4

4

as n→∞.


An area function

Let f (t) = t3 and define g(x) =

∫ x

0f (t) dt. Can we evaluate the integral

in g(x)?

0 x

Dividing the interval [0, x ] into n pieces

gives ∆t =x

nand ti = 0 + i∆t =

ix

n. So

Rn =x

n· x3

n3+

x

n· (2x)3

n3+ · · ·+ x

n· (nx)3

n3

=x4

n4

(13 + 23 + 33 + · · ·+ n3

)=

x4

n4

[12 n(n + 1)

]2=

x4n2(n + 1)2

4n4→ x4

4

as n→∞.


An area function, continued

So

g(x) =x4

4.

This means thatg ′(x) = x3.


An area function, continued

So

g(x) =x4

4.

This means thatg ′(x) = x3.


The area function

Let f be a function which is integrable (i.e., continuous or with finitelymany jump discontinuities) on [a, b]. Define

g(x) =

∫ x

af (t) dt.

I The variable is x ; t is a “dummy” variable that’s integrated over.

I Picture changing x and taking more of less of the region under thecurve.

I Question: What does f tell you about g?


Envisioning the area function

Example

Suppose f (t) is the function graphed below:

x

y

f2 4 6 8 10

Let g(x) =

∫ x

0f (t) dt. What can you say about g?



Example


x

y

g

f2 4 6 8 10

Let g(x) =

∫ x




Example


x

y

g

f2 4 6 8 10

Let g(x) =

∫ x




Example


x

y

g

f2 4 6 8 10

Let g(x) =

∫ x




Example


x

y

g

f2 4 6 8 10

Let g(x) =

∫ x




Example


x

y

g

f2 4 6 8 10

Let g(x) =

∫ x




Example


x

y

g

f2 4 6 8 10

Let g(x) =

∫ x




Example


x

y

g

f2 4 6 8 10

Let g(x) =

∫ x




Example


x

y

g

f2 4 6 8 10

Let g(x) =

∫ x




Example


x

y

g

f2 4 6 8 10

Let g(x) =

∫ x




Example


x

y

g

f2 4 6 8 10

Let g(x) =

∫ x




Example


x

y

g

f2 4 6 8 10

Let g(x) =

∫ x



features of g from f

x

y

g

f2 4 6 8 10

Interval sign monotonicity monotonicity concavity

of f of g of f of g

[0, 2] + ↗ ↗ ^

[2, 4.5] + ↗ ↘ _

[4.5, 6] − ↘ ↘ _

[6, 8] − ↘ ↗ ^

[8, 10] − ↘ → none

We see that g is behaving a lot like an antiderivative of f .


features of g from f

x

y

g

f2 4 6 8 10

Interval sign monotonicity monotonicity concavity

of f of g of f of g

[0, 2] + ↗ ↗ ^

[2, 4.5] + ↗ ↘ _

[4.5, 6] − ↘ ↘ _

[6, 8] − ↘ ↗ ^

[8, 10] − ↘ → none

We see that g is behaving a lot like an antiderivative of f .


Another Big Time Theorem

Theorem (The First Fundamental Theorem of Calculus)

Let f be an integrable function on [a, b] and define

g(x) =

∫ x

af (t) dt.

If f is continuous at x in (a, b), then g is differentiable at x and

g ′(x) = f (x).


Proving the Fundamental Theorem

Proof.

Let h > 0 be given so that x + h < b. We have

g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.

Let Mh be the maximum value of f on [x , x + h], and let mh the minimumvalue of f on [x , x + h]. From §5.2 we have

mh · h ≤

∫ x+h

xf (t) dt

≤ Mh · h

So

mh ≤g(x + h)− g(x)

h≤ Mh.

As h→ 0, both mh and Mh tend to f (x).



Proof.


g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.


mh · h ≤

∫ x+h

xf (t) dt

≤ Mh · h

So

mh ≤g(x + h)− g(x)

h≤ Mh.




Proof.


g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.


mh · h ≤

∫ x+h

xf (t) dt

≤ Mh · h

So

mh ≤g(x + h)− g(x)

h≤ Mh.




Proof.


g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.


mh · h ≤

∫ x+h

xf (t) dt ≤ Mh · h

So

mh ≤g(x + h)− g(x)

h≤ Mh.




Proof.


g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.


mh · h ≤∫ x+h


So

mh ≤g(x + h)− g(x)

h≤ Mh.




Proof.


g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.


mh · h ≤∫ x+h


So

mh ≤g(x + h)− g(x)

h≤ Mh.




Proof.


g(x + h)− g(x)

h=

1

h

∫ x+h

xf (t) dt.


mh · h ≤∫ x+h


So

mh ≤g(x + h)− g(x)

h≤ Mh.



Meet the Mathematician: James Gregory

I Scottish, 1638-1675

I Astronomer and Geometer

I Conceived transcendentalnumbers and found evidencethat π was transcendental

I Proved a geometric versionof 1FTC as a lemma butdidn’t take it further


Meet the Mathematician: Isaac Barrow

I English, 1630-1677

I Professor of Greek, theology,and mathematics atCambridge

I Had a famous student


Meet the Mathematician: Isaac Newton

I English, 1643–1727

I Professor at Cambridge(England)

I Philosophiae NaturalisPrincipia Mathematicapublished 1687


Meet the Mathematician: Gottfried Leibniz

I German, 1646–1716

I Eminent philosopher as wellas mathematician

I Contemporarily disgraced bythe calculus priority dispute


Differentiation and Integration as reverse processes

Putting together 1FTC and 2FTC, we get a beautiful relationship betweenthe two fundamental concepts in calculus.

Theorem (The Fundamental Theorem(s) of Calculus)

I. If f is a continuous function, then

d

dx

∫ x

af (t) dt = f (x)

So the derivative of the integral is the original function.

II. If f is a differentiable function, then∫ b

af ′(x) dx = f (b)− f (a).

So the integral of the derivative of is (an evaluation of) the originalfunction.


Outline





Differentiation of area functions

Example

Let h(x) =

∫ 3x

0t3 dt. What is h′(x)?

Solution (Using 2FTC)

h(x) =t4

4

∣∣∣∣3x0

=1

4(3x)4 = 1

4 · 81x4, so h′(x) = 81x3.


We can think of h as the composition g ◦ k, where g(u) =

∫ u

0t3 dt and

k(x) = 3x. Then h′(x) = g ′(u) · k ′(x), or

h′(x) = g ′(k(x)) · k ′(x) = (k(x))3 · 3 = (3x)3 · 3 = 81x3.



Example

Let h(x) =

∫ 3x



h(x) =t4

4

∣∣∣∣3x0

=1

4(3x)4 = 1

4 · 81x4, so h′(x) = 81x3.



∫ u

0t3 dt and


h′(x) = g ′(k(x)) · k ′(x) = (k(x))3 · 3 = (3x)3 · 3 = 81x3.



Example

Let h(x) =

∫ 3x



h(x) =t4

4

∣∣∣∣3x0

=1

4(3x)4 = 1

4 · 81x4, so h′(x) = 81x3.



∫ u

0t3 dt and

k(x) = 3x.

Then h′(x) = g ′(u) · k ′(x), or

h′(x) = g ′(k(x)) · k ′(x) = (k(x))3 · 3 = (3x)3 · 3 = 81x3.



Example

Let h(x) =

∫ 3x



h(x) =t4

4

∣∣∣∣3x0

=1

4(3x)4 = 1

4 · 81x4, so h′(x) = 81x3.



∫ u

0t3 dt and


h′(x) = g ′(k(x)) · k ′(x) = (k(x))3 · 3 = (3x)3 · 3 = 81x3.


Differentiation of area functions, in general

I by 1FTCd

dx

∫ k(x)

af (t) dt = f (k(x))k ′(x)

I by reversing the order of integration:

d

dx

∫ b

h(x)f (t) dt = − d

dx

∫ h(x)

bf (t) dt = −f (h(x))h′(x)

I by combining the two above:

d

dx

∫ k(x)

h(x)f (t) dt =

d

dx

(∫ k(x)

0f (t) dt +

∫ 0

h(x)f (t) dt

)= f (k(x))k ′(x)− f (h(x))h′(x)


Another Example

Example

Let h(x) =

∫ sin2 x

0(17t2 + 4t − 4) dt. What is h′(x)?

Solution

We have

d

dx

∫ sin2 x

0(17t2 + 4t − 4) dt

=(17(sin2 x)2 + 4(sin2 x)− 4

)· d

dxsin2 x

=(17 sin4 x + 4 sin2 x − 4

)· 2 sin x cos x


Another Example

Example

Let h(x) =

∫ sin2 x

0(17t2 + 4t − 4) dt. What is h′(x)?

Solution

We have

d

dx

∫ sin2 x

0(17t2 + 4t − 4) dt

=(17(sin2 x)2 + 4(sin2 x)− 4

)· d

dxsin2 x

=(17 sin4 x + 4 sin2 x − 4

)· 2 sin x cos x


A Similar Example

Example

Let h(x) =

∫ sin2 x

3(17t2 + 4t − 4) dt. What is h′(x)?

Solution

We have

d

dx

∫ sin2 x

0(17t2 + 4t − 4) dt

=(17(sin2 x)2 + 4(sin2 x)− 4

)· d

dxsin2 x

=(17 sin4 x + 4 sin2 x − 4

)· 2 sin x cos x


A Similar Example

Example

Let h(x) =

∫ sin2 x

3(17t2 + 4t − 4) dt. What is h′(x)?

Solution

We have

d

dx

∫ sin2 x

0(17t2 + 4t − 4) dt

=(17(sin2 x)2 + 4(sin2 x)− 4

)· d

dxsin2 x

=(17 sin4 x + 4 sin2 x − 4

)· 2 sin x cos x


Compare

Question

Why is

d

dx

∫ sin2 x

0(17t2 + 4t − 4) dt =

d

dx

∫ sin2 x

3(17t2 + 4t − 4) dt?

Or, why doesn’t the lower limit appear in the derivative?

Answer

Because∫ sin2 x

0(17t2 + 4t−4) dt =

∫ 3

0(17t2 + 4t−4) dt +

∫ sin2 x

3(17t2 + 4t−4) dt

So the two functions differ by a constant.


Compare

Question

Why is

d

dx

∫ sin2 x

0(17t2 + 4t − 4) dt =

d

dx

∫ sin2 x

3(17t2 + 4t − 4) dt?

Or, why doesn’t the lower limit appear in the derivative?

Answer

Because∫ sin2 x

0(17t2 + 4t−4) dt =

∫ 3

0(17t2 + 4t−4) dt +

∫ sin2 x

3(17t2 + 4t−4) dt

So the two functions differ by a constant.


The Full Nasty

Example

Find the derivative of F (x) =

∫ ex

x3

sin4 t dt.

Solution

d

dx

∫ ex

x3

sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2

Notice here it’s much easier than finding an antiderivative for sin4.


The Full Nasty

Example


∫ ex

x3

sin4 t dt.

Solution

d

dx

∫ ex

x3




The Full Nasty

Example


∫ ex

x3

sin4 t dt.

Solution

d

dx

∫ ex

x3




Why use 1FTC?

Question

Why would we use 1FTC to find the derivative of an integral? It seemslike confusion for its own sake.

Answer

I Some functions are difficult or impossible to integrate in elementaryterms.

I Some functions are naturally defined in terms of other integrals.


Why use 1FTC?

Question


Answer




Why use 1FTC?

Question


Answer




Erf

Here’s a function with a funny name but an important role:

erf(x) =2√π

∫ x

0e−t

2dt.

It turns out erf is the shape of the bell curve. We can’t find erf(x),

explicitly, but we do know its derivative: erf ′(x) =2√π

e−x2.

Example

Findd

dxerf(x2).

Solution

By the chain rule we have

d

dxerf(x2) = erf ′(x2)

d

dxx2 =

2√π

e−(x2)22x =

4√π

xe−x4.


Erf


erf(x) =2√π

∫ x

0e−t

2dt.

It turns out erf is the shape of the bell curve.

We can’t find erf(x),


e−x2.

Example

Findd

dxerf(x2).

Solution


d


d

dxx2 =

2√π

e−(x2)22x =

4√π

xe−x4.


Erf


erf(x) =2√π

∫ x

0e−t

2dt.


explicitly, but we do know its derivative: erf ′(x) =

2√π

e−x2.

Example

Findd

dxerf(x2).

Solution


d


d

dxx2 =

2√π

e−(x2)22x =

4√π

xe−x4.


Erf


erf(x) =2√π

∫ x

0e−t

2dt.



e−x2.

Example

Findd

dxerf(x2).

Solution


d


d

dxx2 =

2√π

e−(x2)22x =

4√π

xe−x4.


Erf


erf(x) =2√π

∫ x

0e−t

2dt.



e−x2.

Example

Findd

dxerf(x2).

Solution


d


d

dxx2 =

2√π

e−(x2)22x =

4√π

xe−x4.


Erf


erf(x) =2√π

∫ x

0e−t

2dt.



e−x2.

Example

Findd

dxerf(x2).

Solution


d


d

dxx2 =

2√π

e−(x2)22x =

4√π

xe−x4.


Other functions defined by integrals

I The future value of an asset:

FV (t) =

∫ ∞t

π(s)e−rs ds

where π(s) is the profitability at time s and r is the discount rate.

I The consumer surplus of a good:

CS(q∗) =

∫ q∗

0(f (q)− p∗) dq

where f (q) is the demand function and p∗ and q∗ the equilibriumprice and quantity.


Surplus by picture

quantity (q)

price (p)

demand f (q)

market revenue

supply

equilibrium

q∗

p∗

consumer surplus

producer surplus


Surplus by picture

quantity (q)

price (p)

demand f (q)

market revenue

supply

equilibrium

q∗

p∗

consumer surplus

producer surplus


Surplus by picture

quantity (q)

price (p)

demand f (q)

market revenue

supply

equilibrium

q∗

p∗

consumer surplus

producer surplus


Surplus by picture

quantity (q)

price (p)

demand f (q)

market revenue

supply

equilibrium

q∗

p∗

consumer surplus

producer surplus


Surplus by picture

quantity (q)

price (p)

demand f (q)

market revenue

supply

equilibrium

q∗

p∗

consumer surplus

producer surplus


Surplus by picture

quantity (q)

price (p)

demand f (q)

market revenue

supply

equilibrium

q∗

p∗

consumer surplus

producer surplus


Surplus by picture

quantity (q)

price (p)

demand f (q)

market revenue

supply

equilibrium

q∗

p∗

consumer surplus

producer surplus


Summary

I Functions defined as integrals can be differentiated using the firstFTC:

d

dx

∫ x

af (t) dt = f (x)

I The two FTCs link the two major processes in calculus: differentiationand integration ∫

F ′(x) dx = F (x) + C

I Follow the calculus wars on twitter: #calcwars


http://twitter.com/#search?q=%23calcwars

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Lesson 27: The Fundamental Theorem of Calculus