The fundamental theorem shows that differentiation and integration are inverse processes.

1. Sec on 5.4
The Fundamental Theorem of
Calculus
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University
May 2, 2011
.

2. Announcements
Today: 5.4
Wednesday 5/4: 5.5
Monday 5/9: Review and
Movie Day!
Thursday 5/12: Final
Exam, 2:00–3:50pm

3. Objectives
State and explain the
Fundamental Theorems of
Calculus
Use the first fundamental
theorem of calculus to find
deriva ves of func ons defined
as integrals.
Compute the average value of
an integrable func on over a
closed interval.

4. Outline
Recall: The Evalua on Theorem a/k/a 2nd FTC
The First Fundamental Theorem of Calculus
Area as a Func on
Statement and proof of 1FTC
Biographies
Differen a on of func ons defined by integrals
“Contrived” examples
Erf
Other applica ons

5. The definite integral as a limit
Defini on
If f is a func on 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 n→∞
i=1
b−a
where ∆x = , and for each i, xi = a + i∆x, and ci is a point in
n
[xi−1 , xi ].

6. The definite integral as a limit
Theorem
If f is con nuous on [a, b] or if f has only finitely many jump
discon nui es, then f is integrable on [a, b]; that is, the definite
∫ b
integral f(x) dx exists and is the same for any choice of ci .
a

7. Big time Theorem
Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F′ for another func on F,
then ∫ b
f(x) dx = F(b) − F(a).
a

8. The Integral as Net Change

9. The Integral as Net Change
Corollary
If v(t) represents the velocity of a par cle moving rec linearly, then
∫ t1
v(t) dt = s(t1 ) − s(t0 ).
t0

10. The Integral as Net Change
Corollary
If MC(x) represents the marginal cost of making x units of a product,
then ∫ x
C(x) = C(0) + MC(q) dq.
0

11. The Integral as Net Change
Corollary
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

12. My first table of integrals
.
∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx
∫ ∫ ∫
xn+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

13. Outline
Recall: The Evalua on Theorem a/k/a 2nd FTC
The First Fundamental Theorem of Calculus
Area as a Func on
Statement and proof of 1FTC
Biographies
Differen a on of func ons defined by integrals
“Contrived” examples
Erf
Other applica ons

14. Area as a Function
Example
∫ x
3
Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x).
0

15. Area as a Function
Example
∫ x
3
Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x).
0
Solu on
x
Dividing the interval [0, x] into n pieces gives ∆t = and
n
ix
ti = 0 + i∆t = .
n
.
0 x

16. Area as a Function
Example
∫ x
3
Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x).
0
Solu on
x x3 x (2x)3 x (nx)3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
.
0 x

17. Area as a Function
Example
∫ x
3
Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x).
0
Solu on
x x3 x (2x)3 x (nx)3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
4 ( )
x
= 4 13 + 23 + 33 + · · · + n3
. n
0 x

18. Area as a Function
Example
∫ x
3
Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x).
0
Solu on
x x3 x (2x)3 x (nx)3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
4 ( )
x x4 [ 1 ]2
= 4 1 + 2 + 3 + · · · + n = 4 2 n(n + 1)
3 3 3 3
. n n
0 x

19. Area as a Function
Example
∫ x
3
Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x).
0
Solu on
x4 n2 (n + 1)2
Rn =
4n4
.
0 x

20. Area as a Function
Example
∫ x
3
Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x).
0
Solu on
x4 n2 (n + 1)2
Rn =
4n4
x4
. So g(x) = lim Rn =
x x→∞ 4
0

21. Area as a Function
Example
∫ x
3
Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x).
0
Solu on
x4 n2 (n + 1)2
Rn =
4n4
x4
. So g(x) = lim Rn = and g′ (x) = x3 .
x x→∞ 4
0

22. The area function in general
Let f be a func on which is integrable (i.e., con nuous or with
finitely many jump discon nui es) 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.
Ques on: What does f tell you about g?

23. Envisioning the area function
Example
Suppose f(t) is the func on y
graphed to the right. Let
∫ x
g(x) = f(t) dt
0
.
x
What can you say about g? 2 4 6 8 10f

24. Envisioning the area function
Example
Suppose f(t) is the func on y
graphed to the right. Let
∫ x
g(x) = f(t) dt
0 g
.
x
What can you say about g? 2 4 6 8 10f

25. Envisioning the area function
Example
Suppose f(t) is the func on y
graphed to the right. Let
∫ x
g(x) = f(t) dt
0 g
.
x
What can you say about g? 2 4 6 8 10f

26. Envisioning the area function
Example
Suppose f(t) is the func on y
graphed to the right. Let
∫ x
g(x) = f(t) dt
0 g
.
x
What can you say about g? 2 4 6 8 10f

27. Envisioning the area function
Example
Suppose f(t) is the func on y
graphed to the right. Let
∫ x
g(x) = f(t) dt
0 g
.
x
What can you say about g? 2 4 6 8 10f

28. Envisioning the area function
Example
Suppose f(t) is the func on y
graphed to the right. Let
∫ x
g(x) = f(t) dt
0 g
.
x
What can you say about g? 2 4 6 8 10f

29. Envisioning the area function
Example
Suppose f(t) is the func on y
graphed to the right. Let
∫ x
g(x) = f(t) dt
0 g
.
x
What can you say about g? 2 4 6 8 10f

30. Envisioning the area function
Example
Suppose f(t) is the func on y
graphed to the right. Let
∫ x
g(x) = f(t) dt
0 g
.
x
What can you say about g? 2 4 6 8 10f

31. Envisioning the area function
Example
Suppose f(t) is the func on y
graphed to the right. Let
∫ x
g(x) = f(t) dt
0 g
.
x
What can you say about g? 2 4 6 8 10f

32. Envisioning the area function
Example
Suppose f(t) is the func on y
graphed to the right. Let
∫ x
g(x) = f(t) dt
0 g
.
x
What can you say about g? 2 4 6 8 10f

33. Envisioning the area function
Example
Suppose f(t) is the func on y
graphed to the right. Let
∫ x
g(x) = f(t) dt
0 g
.
x
What can you say about g? 2 4 6 8 10f

34. Envisioning the area function
Example
Suppose f(t) is the func on y
graphed to the right. Let
∫ x
g(x) = f(t) dt
0 g
.
x
What can you say about g? 2 4 6 8 10f

35. Features of g from f
Interval sign monotonicity monotonicity concavity
y of f of g of f of g
[0, 2] + ↗ ↗ ⌣
g [2, 4.5] + ↗ ↘ ⌢
.
fx
2 4 6 810 [4.5, 6] − ↘ ↘ ⌢
[6, 8] − ↘ ↗ ⌣
[8, 10] − ↘ → none

36. Features of g from f
Interval sign monotonicity monotonicity concavity
y of f of g of f of g
[0, 2] + ↗ ↗ ⌣
g [2, 4.5] + ↗ ↘ ⌢
.
fx
2 4 6 810 [4.5, 6] − ↘ ↘ ⌢
[6, 8] − ↘ ↗ ⌣
[8, 10] − ↘ → none
We see that g is behaving a lot like an an deriva ve of f.

37. Another Big Time Theorem
Theorem (The First Fundamental Theorem of Calculus)
Let f be an integrable func on on [a, b] and define
∫ x
g(x) = f(t) dt.
a
If f is con nuous at x in (a, b), then g is differen able at x and
g′ (x) = f(x).

38. Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
g(x + h) − g(x)
=
h

39. Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
∫
g(x + h) − g(x) 1 x+h
= f(t) dt.
h h x

40. Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
∫
g(x + h) − g(x) 1 x+h
= 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].

41. Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
∫
g(x + h) − g(x) 1 x+h
= 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

42. Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
∫
g(x + h) − g(x) 1 x+h
= 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 ≤ Mh · h
x

43. Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
∫
g(x + h) − g(x) 1 x+h
= 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
mh · h ≤ f(t) dt ≤ Mh · h
x

44. Proving the Fundamental Theorem
Proof.
From §5.2 we have
∫ x+h
mh · h ≤ f(t) dt ≤ Mh · h
x
g(x + h) − g(x)
=⇒ mh ≤ ≤ Mh .
h
As h → 0, both mh and Mh tend to f(x).

45. About Mh and mh
Mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

46. About Mh and mh
Mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

47. About Mh and mh
Since f is con nuous at x, Mh
and x + h → x, we have
lim Mh = f(x)
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

48. About Mh and mh
Since f is con nuous at x,
Mh
and x + h → x, we have
lim Mh = f(x)
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

49. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have Mh
lim Mh = f(x)
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

50. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
Mh
lim Mh = f(x)
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

51. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
Mh
lim Mh = f(x)
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

52. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x) Mh
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

53. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x) Mh
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

54. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0 Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

55. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

56. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

57. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

58. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

59. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

60. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

61. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

62. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

63. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

64. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

65. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

66. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

67. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
xx+h

68. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
xx + h

69. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
xx + h

70. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x +h
x

71. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x+h

72. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
xx+ h

73. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
xx h
+

74. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0 mh
.
x+h
x

75. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
lim mh = f(x) f(x)
h→0
This is not necessarily
true when f is not
con nuous.
.
x

76. About Mh and mh
Since f is con nuous at x, Mh
and x + h → x, we have
lim Mh = f(x)
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

77. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have Mh
lim Mh = f(x)
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

78. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
Mh
lim Mh = f(x)
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

79. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
Mh
lim Mh = f(x)
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

80. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x) Mh
h→0
lim mh = f(x) f(x)
h→0
mh
.
x x+h

81. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0 Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

82. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

83. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

84. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

85. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

86. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

87. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

88. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

89. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

90. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

91. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

92. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

93. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

94. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

95. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

96. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

97. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x x+h

98. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
xx+h

99. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
xx + h

100. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
xx + h

101. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x +h
x

102. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x+h

103. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
xx+ h

104. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
xx h
+

105. About Mh and mh
Since f is con nuous at x,
and x + h → x, we have
lim Mh = f(x)
h→0
Mh
lim mh = f(x) f(x)
h→0
mh
.
x+h
x

106. Meet the Mathematician
James Gregory
Sco sh, 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

107. Meet the Mathematician
Isaac Barrow
English, 1630-1677
Professor of Greek, theology,
and mathema cs at Cambridge
Had a famous student

108. Meet the Mathematician
Isaac Newton
English, 1643–1727
Professor at Cambridge
(England)
invented calculus 1665–66
Tractus de Quadratura
Curvararum published 1704

109. Meet the Mathematician
Gottfried Leibniz
German, 1646–1716
Eminent philosopher as well as
mathema cian
invented calculus 1672–1676
published in papers 1684 and
1686

110. Differentiation and Integration as
reverse processes
Pu ng together 1FTC and 2FTC, we get a beau ful rela onship
between the two fundamental concepts in calculus.
Theorem (The Fundamental Theorem(s) of Calculus)
I. If f is a con nuous func on, then
∫
d x
f(t) dt = f(x)
dx a
So the deriva ve of the integral is the original func on.

111. Differentiation and Integration as
reverse processes
Pu ng together 1FTC and 2FTC, we get a beau ful rela onship
between the two fundamental concepts in calculus.
Theorem (The Fundamental Theorem(s) of Calculus)
II. If f is a differen able func on, then
∫ b
f′ (x) dx = f(b) − f(a).
a
So the integral of the deriva ve of is (an evalua on of) the
original func on.

112. Outline
Recall: The Evalua on Theorem a/k/a 2nd FTC
The First Fundamental Theorem of Calculus
Area as a Func on
Statement and proof of 1FTC
Biographies
Differen a on of func ons defined by integrals
“Contrived” examples
Erf
Other applica ons

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

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

115. Differentiation of area functions
Example
∫ 3x
Let h(x) = t3 dt. What is h′ (x)?
0
Solu on (Using 1FTC)
∫ u
We can think of h as the composi on g k, where g(u) =
◦ t3 dt
0
and k(x) = 3x.

116. Differentiation of area functions
Example
∫ 3x
Let h(x) = t3 dt. What is h′ (x)?
0
Solu on (Using 1FTC)
∫ u
We can think of h as the composi on g k, where g(u) =
◦ t3 dt
0
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 .

117. Differentiation of area functions, in general
by 1FTC ∫
d k(x)
f(t) dt = f(k(x))k′ (x)
dx a
by reversing the order of integra on:
∫ ∫
d b d h(x)
f(t) dt = − f(t) dt = −f(h(x))h′ (x)
dx h(x) dx b
by combining the two above:
∫ (∫ ∫ 0 )
d k(x) d k(x)
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)

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

119. Another Example
Example
∫ sin2 x
Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)?
0
Solu on (2FTC)
sin2 x
17 3 17
h(x) = t + 2t2 − 4t = (sin2 x)3 + 2(sin2 x)2 − 4(sin2 x)
3 3 3
17 6
= sin x + 2 sin4 x − 4 sin2 x
(3 )
′ 17 · 6 5
h (x) = sin x + 2 · 4 sin x − 4 · 2 sin x cos x
3
3

120. Another Example
Example
∫ sin2 x
Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)?
0
Solu on (1FTC)
∫
d sin2 x ( ) d
(17t2 + 4t − 4) dt = 17(sin2 x)2 + 4(sin2 x) − 4 · sin2 x
dx dx
0 ( )
= 17 sin x + 4 sin x − 4 · 2 sin x cos x
4 2

121. A Similar Example
Example
∫ sin2 x
Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)?
3

122. A Similar Example
Example
∫ sin2 x
Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)?
3
Solu on
We have
∫ 2
d sin x ( ) d
(17t2 + 4t − 4) dt = 17(sin2 x)2 + 4(sin2 x) − 4 · sin2 x
dx 3 dx
( )
= 17 sin x + 4 sin x − 4 · 2 sin x cos x
4 2

123. Compare
Ques on
∫ 2 ∫ 2
d sin x d sin x
Why is (17t + 4t − 4) dt =
2
(17t2 + 4t − 4) dt?
dx 0 dx 3
Or, why doesn’t the lower limit appear in the deriva ve?

124. Compare
Ques on
∫ 2 ∫ 2
d sin x d sin x
Why is (17t + 4t − 4) dt =
2
(17t2 + 4t − 4) dt?
dx 0 dx 3
Or, why doesn’t the lower limit appear in the deriva ve?
Answer
∫ sin2 x ∫ 3 ∫ sin2 x
2 2
(17t +4t−4) dt = (17t +4t−4) dt+ (17t2 +4t−4) dt
0 0 3
So the two func ons differ by a constant.

125. The Full Nasty
Example
∫ ex
Find the deriva ve of F(x) = sin4 t dt.
x3

126. The Full Nasty
Example
∫ ex
Find the deriva ve of F(x) = sin4 t dt.
x3
Solu on
∫ ex
d
sin4 t dt = sin4 (ex ) · ex − sin4 (x3 ) · 3x2
dx x3

127. The Full Nasty
Example
∫ ex
Find the deriva ve of F(x) = sin4 t dt.
x3
Solu on
∫ ex
d
sin4 t dt = sin4 (ex ) · ex − sin4 (x3 ) · 3x2
dx x3
No ce here it’s much easier than finding an an deriva ve for sin4 .

128. Why use 1FTC?
Ques on
Why would we use 1FTC to find the deriva ve of an integral? It
seems like confusion for its own sake.

129. Why use 1FTC?
Ques on
Why would we use 1FTC to find the deriva ve of an integral? It
seems like confusion for its own sake.
Answer
Some func ons are difficult or impossible to integrate in
elementary terms.

130. Why use 1FTC?
Ques on
Why would we use 1FTC to find the deriva ve of an integral? It
seems like confusion for its own sake.
Answer
Some func ons are difficult or impossible to integrate in
elementary terms.
Some func ons are naturally defined in terms of other
integrals.

131. Erf
∫ x
2
e−t dt
2
erf(x) = √
π 0

132. Erf
∫ x
2
e−t dt
2
erf(x) = √
π 0
erf measures area the bell curve. erf(x)
.
x

133. Erf
∫ x
2
e−t dt
2
erf(x) = √
π 0
erf measures area the bell curve. erf(x)
We can’t find erf(x), explicitly,
but we do know its deriva ve: .
x
erf′ (x) =

134. Erf
∫ x
2
e−t dt
2
erf(x) = √
π 0
erf measures area the bell curve. erf(x)
We can’t find erf(x), explicitly,
but we do know its deriva ve: .
2 x
erf′ (x) = √ e−x .
2
π

135. Example of erf
Example
d
Find erf(x2 ).
dx

136. Example of erf
Example
d
Find erf(x2 ).
dx
Solu on
By the chain rule we have
d d 2 4
erf(x2 ) = erf′ (x2 ) x2 = √ e−(x ) 2x = √ xe−x .
2 2 4
dx dx π π

137. 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 me 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 func on and p∗ and q∗ the
equilibrium price and quan ty.

138. Surplus by picture
price (p)
.
quan ty (q)

139. Surplus by picture
price (p)
demand f(q)
.
quan ty (q)

140. Surplus by picture
price (p)
supply
demand f(q)
.
quan ty (q)

141. Surplus by picture
price (p)
supply
p∗ equilibrium
demand f(q)
.
q∗ quan ty (q)

142. Surplus by picture
price (p)
supply
p∗ equilibrium
market revenue
demand f(q)
.
q∗ quan ty (q)

143. Surplus by picture
consumer surplus
price (p)
supply
p∗ equilibrium
market revenue
demand f(q)
.
q∗ quan ty (q)

144. Surplus by picture
consumer surplus
price (p)
producer surplus
supply
p∗ equilibrium
demand f(q)
.
q∗ quan ty (q)

145. Summary
Func ons defined as integrals can be differen ated using the
first FTC: ∫
d x
f(t) dt = f(x)
dx a
The two FTCs link the two major processes in calculus:
differen a on and integra on
∫
F′ (x) dx = F(x) + C