Lesson 27: The Fundamental Theorem of Calculus

Section 5.4
The Fundamental Theorem of Calculus

V63.0121.002.2010Su, Calculus I

New York University

June 21, 2010

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Announcements

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

Objectives

State and explain the
Fundemental Theorems of
Calculus
Use the first fundamental
theorem of calculus to find
derivatives of functions
defined as integrals.
Compute the average value
of an integrable function
over a closed interval.


Outline

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

The First Fundamental Theorem of Calculus
The Area Function
Statement and proof of 1FTC
Biographies

Diﬀerentiation of functions deﬁned by integrals
“Contrived” examples
Erf
Other 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 b
is the number
b n
f (x) dx = lim f (ci ) ∆x
a ∆x→0
i=1


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
f (x) dx = F (b) − F (a).
a


The Integral as Total Change

Another way to state this theorem is:
b
F (x) dx = F (b) − F (a),
a

or the integral of a derivative along an interval is the total change between
the sides of that interval. This has many ramiﬁcations:



b
F (x) dx = F (b) − F (a),
a


Theorem
If v (t) represents the velocity of a particle moving rectilinearly, then
t1
v (t) dt = s(t1 ) − s(t0 ).
t0



b
F (x) dx = F (b) − F (a),
a


Theorem
If MC (x) represents the marginal cost of making x units of a product, then
x
C (x) = C (0) + MC (q) dq.
0



b
F (x) dx = F (b) − F (a),
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
x
m(x) = ρ(s) ds.
0


My ﬁrst table of integrals

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

x n+1
x n dx = + C (n = −1) cf (x) dx = c f (x) dx
n+1
1
e x dx = e x + C dx = ln |x| + C
x
ax
sin x dx = − cos x + C ax dx = +C
ln a

cos x dx = sin x + C csc2 x dx = − cot x + C

sec2 x dx = tan x + C csc x cot x dx = − csc x + C
1
sec x tan x dx = sec x + C √ dx = arcsin x + C
1 − x2
1
dx = arctan x + C
1 + x2


Outline


The Area Function
Biographies

Erf
Other applications


An area function
x
Let f (t) = t 3 and deﬁne g (x) = f (t) dt. Can we evaluate the integral
0
in g (x)?

0 x


An area function
x
Let f (t) = t 3 and deﬁne g (x) = f (t) dt. Can we evaluate the integral
0
in g (x)?
Dividing the interval [0, x] into n pieces
x ix
gives ∆t = and ti = 0 + i∆t = . So
n n
x x 3 x (2x)3 x (nx)3
Rn = · 3+ · 3
+ ··· + ·
n n n n n n3
x4
= 4 13 + 23 + 33 + · · · + n3
n
x4 1 2
= 4 2 n(n + 1)
n
0 x
x 4 n2 (n + 1)2 x4
= 4
→
4n 4
as n → ∞.

An area function, continued

So
x4
g (x) = .
4


An area function, continued

So
x4
g (x) = .
4
This means that
g (x) = x 3 .


The area function

Let f be a function which is integrable (i.e., continuous or with ﬁnitely
many jump discontinuities) on [a, b]. Deﬁne
x
g (x) = f (t) dt.
a

The variable is x; t is a “dummy” variable that’s integrated over.
Picture changing x and taking more of less of the region under the
curve.
Question: What does f tell you about g ?


Envisioning the area function

Example
Suppose f (t) is the function graphed below:
y

x
2 4 6 8 10f

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


Envisioning the area function

Example
Suppose f (t) is the function graphed below:
y

g
x
2 4 6 8 10f

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


features of g from f

y
Interval sign monotonicity monotonicity concavity
of f of g of f of g
g
[0, 2] +
x
2 4 6 8 10f [2, 4.5] +
[4.5, 6] −
[6, 8] −
[8, 10] − → none


features of g from f

y
Interval sign monotonicity monotonicity concavity
of f of g of f of g
g
[0, 2] +
x
2 4 6 8 10f [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
x
g (x) = f (t) dt.
a

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



Proof.
x+h
g (x + h) − g (x) 1
= f (t) dt.
h h x



Proof.
x+h
g (x + h) − g (x) 1
= f (t) dt.
h h x

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



Proof.
x+h
g (x + h) − g (x) 1
= f (t) dt.
h h x

x+h
f (t) dt ≤ Mh · h
x



Proof.
x+h
g (x + h) − g (x) 1
= f (t) dt.
h h x

x+h
mh · h ≤ f (t) dt ≤ Mh · h
x



Proof.
x+h
g (x + h) − g (x) 1
= f (t) dt.
h h x

x+h
mh · h ≤ f (t) dt ≤ Mh · h
x

So
g (x + h) − g (x)
mh ≤ ≤ Mh .
h



Proof.
x+h
g (x + h) − g (x) 1
= f (t) dt.
h h x

x+h
mh · h ≤ f (t) dt ≤ Mh · h
x

So
g (x + h) − g (x)
mh ≤ ≤ Mh .
h
As h → 0, both mh and Mh tend to f (x).


Meet the Mathematician: James Gregory

Scottish, 1638-1675
Astronomer and Geometer
Conceived transcendental
numbers and found evidence
that π was transcendental
Proved a geometric version
of 1FTC as a lemma but
didn’t take it further


Meet the Mathematician: Isaac Barrow

English, 1630-1677
Professor of Greek, theology,
and mathematics at
Cambridge
Had a famous student


Meet the Mathematician: Isaac Newton

English, 1643–1727
Professor at Cambridge
(England)
Philosophiae Naturalis
Principia Mathematica
published 1687


Meet the Mathematician: Gottfried Leibniz

German, 1646–1716
Eminent philosopher as well
as mathematician
Contemporarily disgraced by
the calculus priority dispute


Diﬀerentiation and Integration as reverse processes
Putting together 1FTC and 2FTC, we get a beautiful relationship between
the two fundamental concepts in calculus.
Theorem (The Fundamental Theorem(s) of Calculus)

I. If f is a continuous function, then
x
d
f (t) dt = f (x)
dx a

So the derivative of the integral is the original function.
II. If f is a diﬀerentiable function, then
b
f (x) dx = f (b) − f (a).
a

So the integral of the derivative of is (an evaluation of) the original
function.

Outline


The Area Function
Biographies

Erf
Other applications


Diﬀerentiation of area functions

Example
3x
Let h(x) = t 3 dt. What is h (x)?
0



Example
3x
0

Solution (Using 2FTC)
3x
t4 1
h(x) = = (3x)4 = 1
4 · 81x 4 , so h (x) = 81x 3 .
4 0 4



Example
3x
0

3x
t4 1
h(x) = = (3x)4 = 1
4 · 81x 4 , so h (x) = 81x 3 .
4 0 4

u
We can think of h as the composition g ◦ k, where g (u) = t 3 dt and
0
k(x) = 3x.



Example
3x
0

3x
t4 1
h(x) = = (3x)4 = 1
4 · 81x 4 , so h (x) = 81x 3 .
4 0 4

u
We can think of h as the composition g ◦ k, where g (u) = t 3 dt and
0
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 = 81x 3 .


Diﬀerentiation of area functions, in general

by 1FTC
k(x)
d
f (t) dt = f (k(x))k (x)
dx a
by reversing the order of integration:
b h(x)
d d
f (t) dt = − f (t) dt = −f (h(x))h (x)
dx h(x) dx b

by combining the two above:

k(x) k(x) 0
d d
f (t) dt = f (t) dt + f (t) dt
dx h(x) dx 0 h(x)

= f (k(x))k (x) − f (h(x))h (x)


Another Example

Example
sin2 x
Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)?
0


Another Example

Example
sin2 x
0

Solution
We have
sin2 x
d
(17t 2 + 4t − 4) dt
dx 0
d
= 17(sin2 x)2 + 4(sin2 x) − 4 · sin2 x
dx
= 17 sin4 x + 4 sin2 x − 4 · 2 sin x cos x


A Similar Example

Example
sin2 x
3


A Similar Example

Example
sin2 x
3

Solution
We have
sin2 x
d
(17t 2 + 4t − 4) dt
dx 0
d
= 17(sin2 x)2 + 4(sin2 x) − 4 · sin2 x
dx
= 17 sin4 x + 4 sin2 x − 4 · 2 sin x cos x


Compare

Question
Why is
sin2 x sin2 x
d 2 d
(17t + 4t − 4) dt = (17t 2 + 4t − 4) dt?
dx 0 dx 3

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


Compare

Question
Why is
sin2 x sin2 x
d 2 d
(17t + 4t − 4) dt = (17t 2 + 4t − 4) dt?
dx 0 dx 3

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

Answer
Because
sin2 x 3 sin2 x
2 2
(17t + 4t − 4) dt = (17t + 4t − 4) dt + (17t 2 + 4t − 4) dt
0 0 3

So the two functions diﬀer by a constant.


The Full Nasty

Example
ex
Find the derivative of F (x) = sin4 t dt.
x3


The Full Nasty

Example
ex
x3

Solution

ex
d
sin4 t dt = sin4 (e x ) · e x − sin4 (x 3 ) · 3x 2
dx x3


The Full Nasty

Example
ex
x3

Solution

ex
d
sin4 t dt = sin4 (e x ) · e x − sin4 (x 3 ) · 3x 2
dx x3

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


Why use 1FTC?

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


Why use 1FTC?

Question

Answer

Some functions are diﬃcult or impossible to integrate in elementary
terms.


Why use 1FTC?

Question

Answer

Some functions are diﬃcult or impossible to integrate in elementary
terms.
Some functions are naturally deﬁned in terms of other integrals.


Erf
Here’s a function with a funny name but an important role:
x
2 2
erf(x) = √ e −t dt.
π 0


Erf
x
2 2
π 0
It turns out erf is the shape of the bell curve.


Erf
x
2 2
π 0
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) =


Erf
x
2 2
π 0
2 2
explicitly, but we do know its derivative: erf (x) = √ e −x .
π


Erf
x
2 2
π 0
2 2
π
Example
d
Find erf(x 2 ).
dx


Erf
x
2 2
π 0
2 2
π
Example
d
Find erf(x 2 ).
dx

Solution
By the chain rule we have
d d 2 2 2 4 4
erf(x 2 ) = erf (x 2 ) x 2 = √ e −(x ) 2x = √ xe −x .
dx dx π π


Other functions deﬁned by integrals

The future value of an asset:
∞
FV (t) = π(s)e −rs ds
t

where π(s) is the proﬁtability at time s and r is the discount rate.
The consumer surplus of a good:
q∗
CS(q ∗ ) = (f (q) − p ∗ ) dq
0

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


Surplus by picture

price (p)

quantity (q)


Surplus by picture

price (p)

demand f (q)

quantity (q)


Surplus by picture

price (p)

supply

demand f (q)

quantity (q)


Surplus by picture

price (p)

supply

p∗ equilibrium

demand f (q)

q∗ quantity (q)


Surplus by picture

price (p)

supply

p∗ equilibrium

market revenue

demand f (q)

q∗ quantity (q)


Surplus by picture

consumer surplus
price (p)

supply

p∗ equilibrium

market revenue

demand f (q)

q∗ quantity (q)


Surplus by picture

consumer surplus
price (p)
producer surplus
supply

p∗ equilibrium

demand f (q)

q∗ quantity (q)


Summary

Functions defined as integrals can be differentiated using the first
FTC: x
d
f (t) dt = f (x)
dx a
The two FTCs link the two major processes in calculus: differentiation
and integration
F (x) dx = F (x) + C

Follow the calculus wars on twitter: #calcwars


Lesson 27: The Fundamental Theorem of Calculus

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