This document contains lecture notes on the Fundamental Theorem of Calculus. It begins with announcements about the final exam date and current grade distribution. The outline then reviews the Evaluation Theorem and introduces the First and Second Fundamental Theorems of Calculus. It provides examples of how the integral can represent total change in concepts like distance, cost, and mass. Biographies are also included of mathematicians like Gregory, Barrow, Newton, and Leibniz who contributed to the development of calculus.

1. Section 5.4
The Fundamental Theorem of Calculus
V63.0121.027, Calculus I
December 8, 2009
Announcements
Final Exam: Friday 12/18, 2:00-3:50pm, Tisch UC50
. . . . . .

2. Redemption policies
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
[new!] If your final exam score beats your midterm score,
we will re-weight it by 50% and make the midterm 15%
. . . . . .

3. 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
. . . . . .

4. 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
. . . . . .

5. 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
. . . . . .

6. 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:
. . . . . .

7. 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:
Theorem
If v(t) represents the velocity of a particle moving rectilinearly,
then ∫ t1
v(t) dt = s(t1 ) − s(t0 ).
t0
. . . . . .

8. 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:
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
. . . . . .

9. 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:
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
. . . . . .

10. My first table of integrals
∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx
∫ ∫ ∫
x n +1
xn dx = + C (n ̸= −1) cf(x) dx = c f(x) dx
n+1 ∫
∫
1
ex dx = ex + 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
. . . . . .

11. 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
. . . . . .

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

13. An area function
∫ x
3
Let f(t) = t and define g(x) = f(t) dt. Can we evaluate the
0
integral in g(x)?
Dividing the interval [0, x] into n pieces
x ix
gives ∆t = and ti = 0 + i∆t = . So
n n
x x3 x (2x)3 x (nx)3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
x4 ( 3 )
= 4 1 + 2 3 + 3 3 + · · · + n3
n
x4 [ 1 ]2
= 4 2 n(n + 1)
. n
0
. x
.
x4 n2 (n + 1)2 x4
= →
4n4 4
as n → ∞.
. . . . . .

14. An area function, continued
So
x4
g(x) = .
4
. . . . . .

15. An area function, continued
So
x4
g(x) = .
4
This means that
g ′ (x ) = x 3 .
. . . . . .

16. 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?
. . . . . .

17. Envisioning the area function
Example
Suppose f(t) is the function graphed below
v
. .
. . . . .
t
.0 t
.1 c
. t
.2 t t
.3 .
.
∫ x
Let g(x) = f(t) dt. What can you say about g?
t0
. . . . . .

18. features of g from f
Interval sign monotonicity monotonicity concavity
of f of g of f of g
[ t0 , t 1 ] + ↗ ↗ ⌣
[t1 , c] + ↗ ↘ ⌢
[c, t2 ] − ↘ ↘ ⌢
[ t2 , t 3 ] − ↘ ↗ ⌣
[t3 , ∞) − ↘ → none
. . . . . .

19. features of g from f
Interval sign monotonicity monotonicity concavity
of f of g of f of g
[ t0 , t 1 ] + ↗ ↗ ⌣
[t1 , c] + ↗ ↘ ⌢
[c, t2 ] − ↘ ↘ ⌢
[ t2 , t 3 ] − ↘ ↗ ⌣
[t3 , ∞) − ↘ → none
We see that g is behaving a lot like an antiderivative of f.
. . . . . .

20. 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).
. . . . . .

21. Proof.
Let h > 0 be given so that x + h < b. We have
g(x + h) − g(x)
=
h
. . . . . .

22. Proof.
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
. . . . . .

23. Proof.
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
f(t) dt
x
. . . . . .

24. Proof.
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
f(t) dt ≤ Mh · h
x
. . . . . .

25. Proof.
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
. . . . . .

26. Proof.
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
. . . . . .

27. Proof.
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).
. . . . . .

28. 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
. . . . . .

29. Meet the Mathematician: Isaac Barrow
English, 1630-1677
Professor of Greek,
theology, and
mathematics at
Cambridge
Had a famous student
. . . . . .

30. Meet the Mathematician: Isaac Newton
English, 1643–1727
Professor at Cambridge
(England)
Philosophiae Naturalis
Principia Mathematica
published 1687
. . . . . .

31. Meet the Mathematician: Gottfried Leibniz
German, 1646–1716
Eminent philosopher as
well as mathematician
Contemporarily
disgraced by the
calculus priority dispute
. . . . . .

32. Differentiation and Integration as reverse processes
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
. . . . . .

33. Differentiation and Integration as reverse processes
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
. . . . . .

34. 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
. . . . . .

35. Differentiation of area functions
Example
∫ 3x
Let h(x) = t3 dt. What is h′ (x)?
0
. . . . . .

36. Differentiation of area functions
Example
∫ 3x
Let h(x) = t3 dt. What is h′ (x)?
0
Solution (Using 2FTC)
3x
t4 1
h(x) = = (3x)4 = 1
4 · 81x4 , so h′ (x) = 81x3 .
4 0 4
. . . . . .

37. Differentiation of area functions
Example
∫ 3x
Let h(x) = t3 dt. What is h′ (x)?
0
Solution (Using 2FTC)
3x
t4 1
h(x) = = (3x)4 = 1
4 · 81x4 , so h′ (x) = 81x3 .
4 0 4
Solution (Using 1FTC) ∫ u
We can think of h as the composition g k, where g(u) = ◦ t3 dt
0
and k(x) = 3x.
. . . . . .

38. Differentiation of area functions
Example
∫ 3x
Let h(x) = t3 dt. What is h′ (x)?
0
Solution (Using 2FTC)
3x
t4 1
h(x) = = (3x)4 = 1
4 · 81x4 , so h′ (x) = 81x3 .
4 0 4
Solution (Using 1FTC) ∫ u
We can think of h as the composition g k, where g(u) = ◦ t3 dt
0
and k(x) = 3x. Then
h′ (x) = g′ (k(x))k′ (x) = (k(x))3 · 3 = (3x)3 · 3 = 81x3 .
. . . . . .

39. 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)
. . . . . .

40. Example
∫ sin2 x
Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)?
0
. . . . . .

41. Example
∫ sin2 x
Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)?
0
Solution
We have
∫ sin2 x
d
(17t2 + 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
. . . . . .

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

43. Example
∫ ex
Find the derivative of F(x) = sin4 t dt.
x3
Solution
∫ ex
d
sin4 t dt = sin4 (ex ) · ex − sin4 (x3 ) · 3x2
dx x3
. . . . . .

44. Example
∫ ex
Find the derivative of F(x) = sin4 t dt.
x3
Solution
∫ ex
d
sin4 t dt = sin4 (ex ) · ex − sin4 (x3 ) · 3x2
dx x3
Notice here it’s much easier than finding an antiderivative for
sin4 .
. . . . . .

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

46. Erf
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.
. . . . . .

47. Erf
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),
explicitly, but we do know its derivative.
erf′ (x) =
. . . . . .

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

49. Erf
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),
explicitly, but we do know its derivative.
2 2
erf′ (x) = √ e−x .
π
Example
d
Find erf(x2 ).
dx
. . . . . .

50. Erf
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),
explicitly, but we do know its derivative.
2 2
erf′ (x) = √ e−x .
π
Example
d
Find erf(x2 ).
dx
Solution
By the chain rule we have
d d 2 2 2 4 4
erf(x2 ) = erf′ (x2 ) x2 = √ e−(x ) 2x = √ xe−x .
dx dx π π
. . . . . .

51. Other functions defined by integrals
The future value of an asset:
∫ ∞
FV(t) = π(τ )e−rτ 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.
. . . . . .

52. Surplus by picture
c
. onsumer surplus
p
. rice (p)
s
. upply
.∗ .
p . . quilibrium
e
. emand f(q)
d
. .
.∗
q q
. uantity (q)
. . . . . .