Here are the key points about g given f: - g represents the area under the curve of f over successive intervals of the x-axis - As x increases over an interval, g will increase if f is positive over that interval and decrease if f is negative - The concavity (convexity or concavity) of g will match the concavity of f over each interval In summary, the area function g, as defined by the integral of f, will have properties that correspond directly to the sign and concavity of f over successive intervals of integration.

1. Section 5.4
The Fundamental Theorem of Calculus
V63.0121.002.2010Su, Calculus I
New York University
June 21, 2010
Announcements

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

3. 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 3 / 33

4. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 4 / 33

5. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 5 / 33

6. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 6 / 33

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:
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 7 / 33

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 v (t) represents the velocity of a particle moving rectilinearly, then
t1
v (t) dt = s(t1 ) − s(t0 ).
t0
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 7 / 33

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 MC (x) represents the marginal cost of making x units of a product, then
x
C (x) = C (0) + MC (q) dq.
0
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 7 / 33

10. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 7 / 33

11. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 8 / 33

12. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 9 / 33

13. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 10 / 33

14. 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 → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 10 / 33

15. An area function, continued
So
x4
g (x) = .
4
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 11 / 33

16. An area function, continued
So
x4
g (x) = .
4
This means that
g (x) = x 3 .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 11 / 33

17. 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 ?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 12 / 33

18. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33

19. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33

20. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33

21. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33

22. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33

23. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33

24. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33

25. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33

26. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33

27. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33

28. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33

29. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33

30. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 14 / 33

31. 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 .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 14 / 33

32. 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).
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 15 / 33

33. Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
g (x + h) − g (x)
=
h
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33

34. Proving the Fundamental Theorem
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33

35. Proving the Fundamental Theorem
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 let mh the minimum
value of f on [x, x + h]. From §5.2 we have
x+h
f (t) dt
x
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33

36. Proving the Fundamental Theorem
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 let mh the minimum
value of f on [x, x + h]. From §5.2 we have
x+h
f (t) dt ≤ Mh · h
x
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33

37. Proving the Fundamental Theorem
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 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33

38. Proving the Fundamental Theorem
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 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
So
g (x + h) − g (x)
mh ≤ ≤ Mh .
h
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33

39. Proving the Fundamental Theorem
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 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
So
g (x + h) − g (x)
mh ≤ ≤ Mh .
h
As h → 0, both mh and Mh tend to f (x).
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33

40. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 17 / 33

41. Meet the Mathematician: Isaac Barrow
English, 1630-1677
Professor of Greek, theology,
and mathematics at
Cambridge
Had a famous student
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 18 / 33

42. Meet the Mathematician: Isaac Newton
English, 1643–1727
Professor at Cambridge
(England)
Philosophiae Naturalis
Principia Mathematica
published 1687
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 19 / 33

43. Meet the Mathematician: Gottfried Leibniz
German, 1646–1716
Eminent philosopher as well
as mathematician
Contemporarily disgraced by
the calculus priority dispute
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 20 / 33

44. 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 21 / 33

45. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 22 / 33

46. Differentiation of area functions
Example
3x
Let h(x) = t 3 dt. What is h (x)?
0
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 23 / 33

47. Differentiation of area functions
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 23 / 33

48. Differentiation of area functions
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 23 / 33

49. Differentiation of area functions
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 (u) · k (x), or
h (x) = g (k(x)) · k (x) = (k(x))3 · 3 = (3x)3 · 3 = 81x 3 .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 23 / 33

50. 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)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 24 / 33

51. Another Example
Example
sin2 x
Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)?
0
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 25 / 33

52. Another Example
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 25 / 33

53. A Similar Example
Example
sin2 x
Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)?
3
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 26 / 33

54. A Similar Example
Example
sin2 x
Let h(x) = (17t 2 + 4t − 4) dt. What is h (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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 26 / 33

55. 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?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 27 / 33

56. 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 27 / 33

57. The Full Nasty
Example
ex
Find the derivative of F (x) = sin4 t dt.
x3
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 28 / 33

58. The Full Nasty
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 28 / 33

59. The Full Nasty
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 .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 28 / 33

60. Why use 1FTC?
Question
Why would we use 1FTC to find the derivative of an integral? It seems
like confusion for its own sake.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 29 / 33

61. Why use 1FTC?
Question
Why would we use 1FTC to find the derivative of an integral? It seems
like confusion for its own sake.
Answer
Some functions are difficult or impossible to integrate in elementary
terms.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 29 / 33

62. Why use 1FTC?
Question
Why would we use 1FTC to find the derivative of an integral? It seems
like confusion for its own sake.
Answer
Some functions are difficult or impossible to integrate in elementary
terms.
Some functions are naturally defined in terms of other integrals.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 29 / 33

63. Erf
Here’s a function with a funny name but an important role:
x
2 2
erf(x) = √ e −t dt.
π 0
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33

64. 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33

65. 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) =
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33

66. 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),
2 2
explicitly, but we do know its derivative: erf (x) = √ e −x .
π
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33

67. 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),
2 2
explicitly, but we do know its derivative: erf (x) = √ e −x .
π
Example
d
Find erf(x 2 ).
dx
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33

68. 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),
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33

69. 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 31 / 33

70. Surplus by picture
price (p)
quantity (q)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33

71. Surplus by picture
price (p)
demand f (q)
quantity (q)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33

72. Surplus by picture
price (p)
supply
demand f (q)
quantity (q)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33

73. Surplus by picture
price (p)
supply
p∗ equilibrium
demand f (q)
q∗ quantity (q)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33

74. Surplus by picture
price (p)
supply
p∗ equilibrium
market revenue
demand f (q)
q∗ quantity (q)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33

75. Surplus by picture
consumer surplus
price (p)
supply
p∗ equilibrium
market revenue
demand f (q)
q∗ quantity (q)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33

76. Surplus by picture
consumer surplus
price (p)
producer surplus
supply
p∗ equilibrium
demand f (q)
q∗ quantity (q)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33

77. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 33 / 33