1. Section 5.4
The Fundamental Theorem of Calculus
V63.0121.034, Calculus I
December 7, 2009
Announcements
Final Exam: Friday 12/18, 2:00-3:50pm, Tisch UC50
. . . . . .

2. Redemption policies
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[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)
. . . . . .