Lesson 25: The Fundamental Theorem of Calculus (handout)

V63.0121.006/016, Calculus I 5.4 : The Fundamental Theorem of Calculus
Section April 22, 2010

Section 5.4 Notes

The Fundamental Theorem of Calculus

V63.0121.006/016, Calculus I

New York University

April 22, 2010

Announcements

April 29: Movie Day
April 30: Quiz 5 on §§5.1–5.4
Monday, May 10, 12:00noon Final Exam

Announcements
Notes

April 29: Movie Day
April 30: Quiz 5 on
§§5.1–5.4
Monday, May 10, 12:00noon
Final Exam

V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 2 / 31

Resurrection policies
Notes

Current distribution of grade: 40% ﬁnal, 25% midterm, 15% quizzes,
10% written HW, 10% WebAssign
Remember we drop the lowest quiz, lowest written HW, and 5 lowest
WebAssign-ments
If your ﬁnal exam score beats your midterm score, we will re-weight it
by 50% and make the midterm 15%


1


Objectives
Notes

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
Notes

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
Notes

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


2


Notes

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
Notes

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:

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


Notes

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


3


Notes

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
Notes

[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
x x
1
e dx = e + 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
Notes


The Area Function
Biographies

Erf
Other applications


4


An area function
Notes
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 = · + · + ··· + ·
n n3 n n3 n n3
x4 3
= 4 1 + 23 + 33 + · · · + n 3
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
Notes

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


The area function
Notes

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 ?


5


Envisioning the area function
Notes
Example
Suppose f (t) is the function graphed below
v

t0 t1 c t2 t3 t

x
Let g (x) = f (t) dt. What can you say about g ?
t0


features of g from f
Notes

Interval sign monotonicity monotonicity concavity
of f of g of f of g
[t0 , t1 ] +
[t1 , c] +
[c, t2 ] −
[t2 , t3 ] −
[t3 , ∞) − → none
We see that g is behaving a lot like an antiderivative of f .


Notes

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).


6


Proof. Notes
Let h > 0 be given so that x + h < b. We have
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 mh the minimum
value of f on [x, x + h]. From §5.2 we have
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
Notes

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
Notes

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


7


Meet the Mathematician: Isaac Newton
Notes

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


Meet the Mathematician: Gottfried Leibniz
Notes

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


Diﬀerentiation and Integration as reverse processes
Notes

Putting together 1FTC and 2FTC, we get a beautiful relationship between
the two fundamental concepts in calculus.
x
d
f (t) dt = f (x)
dx a

b
F (x) dx = F (b) − F (a).
a


8


Outline
Notes


The Area Function
Biographies

Erf
Other applications


Diﬀerentiation of area functions
Notes
Example
3x
Let h(x) = t 3 dt. What is h (x)?
0

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

Solution (Using 1FTC)
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 (k(x))k (x) = (k(x))3 · 3 = (3x)3 · 3 = 81x 3 .


Diﬀerentiation of area functions, in general
Notes

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)


9


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


Notes

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


Erf
Notes
Here’s a function with a funny name but an important role:
x
2 2
erf(x) = √ e −t dt.
π 0
It turns out erf is the shape of the bell curve. We can’t ﬁnd erf(x),
2 2
explicitly, but we do know its derivative: erf (x) = √ e −x .
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


10


Other functions defined by integrals
Notes

The future value of an asset:
∞
FV (t) = π(τ )e −r τ dτ
t

where π(τ ) is the profitability at time τ 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
Notes
consumer surplus
price (p)

supply

p∗ equilibrium

demand f (q)

q∗ quantity (q)


Summary
Notes

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


11

Lesson 25: The Fundamental Theorem of Calculus (handout)

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