The document is a lecture note on the fundamental theorem of calculus from a Calculus I class at New York University. It provides announcements about upcoming exams and assignments. It then outlines the key topics to be covered, including the first fundamental theorem of calculus and how to differentiate functions defined by integrals. Examples are provided to illustrate using integrals to find the area under a curve and how this relates to the derivative of the area function.
1. Section 5.4
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
. . . . . . .
2. Announcements
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 April 22, 2010 2 / 34
3. Resurrection 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
If your final exam score beats your midterm score, we will
re-weight it by 50% and make the midterm 15%
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 3 / 34
4. Resurrection 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
If your final exam score beats your midterm score, we will
re-weight it by 50% and make the midterm 15%
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 3 / 34
5. 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.
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 4 / 34
6. 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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 5 / 34
7. 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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 6 / 34
8. 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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 7 / 34
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:
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 8 / 34
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 v(t) represents the velocity of a particle moving rectilinearly, then
∫ t1
v(t) dt = s(t1 ) − s(t0 ).
t0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 8 / 34
11. 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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 8 / 34
12. 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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 8 / 34
13. 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
∫ ∫
2
sec 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
. . . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 9 / 34
14. 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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 10 / 34
15. An area function
∫ x
Let f(t) = t3 and define g(x) = f(t) dt. Can we evaluate the integral
0
in g(x)?
.
0
. x
.
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 11 / 34
16. An area function
∫ x
Let f(t) = t3 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 x3 x (2x)3 x (nx)3
Rn = · 3+ · + ··· + ·
n n n n3 n n3
x4 ( )
= 4 13 + 23 + 33 + · · · + n3
n
x4 [ ]2
= 4 1 n(n + 1)
. n 2
0
. x
.
x4 n2 (n + 1)2 x4
= →
4n4 4
as n → ∞.
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 11 / 34
17. An area function, continued
So
x4
g(x) = .
4
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 12 / 34
18. An area function, continued
So
x4
g(x) = .
4
This means that
g′ (x) = x3 .
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 12 / 34
19. 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?
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 13 / 34
20. Envisioning the area function
Example
Suppose f(t) is the function graphed below:
..
y
.
. . . . . . .
. x
.
2
. 4
. 6
.. 8 1f
. . 0.
∫ x
Let g(x) = f(t) dt. What can you say about g?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
21. Envisioning the area function
Example
Suppose f(t) is the function graphed below:
..
y
.
g
.
. . . . . . .
. x
.
2
. 4
. 6
.. 8 1f
. . 0.
∫ x
Let g(x) = f(t) dt. What can you say about g?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
22. Envisioning the area function
Example
Suppose f(t) is the function graphed below:
..
y
.
g
.
. . . . . . .
. x
.
2
. 4
. 6
.. 8 1f
. . 0.
∫ x
Let g(x) = f(t) dt. What can you say about g?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
23. Envisioning the area function
Example
Suppose f(t) is the function graphed below:
..
y
.
g
.
. . . . . . .
. x
.
2
. 4
. 6
.. 8 1f
. . 0.
∫ x
Let g(x) = f(t) dt. What can you say about g?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
24. Envisioning the area function
Example
Suppose f(t) is the function graphed below:
..
y
.
g
.
. . . . . . .
. x
.
2
. 4
. 6
.. 8 1f
. . 0.
∫ x
Let g(x) = f(t) dt. What can you say about g?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
25. Envisioning the area function
Example
Suppose f(t) is the function graphed below:
..
y
.
g
.
. . . . . . .
. x
.
2
. 4
. 6
.. 8 1f
. . 0.
∫ x
Let g(x) = f(t) dt. What can you say about g?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
26. Envisioning the area function
Example
Suppose f(t) is the function graphed below:
..
y
.
g
.
. . . . . . .
. x
.
2
. 4
. 6
.. 8 1f
. . 0.
∫ x
Let g(x) = f(t) dt. What can you say about g?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
27. Envisioning the area function
Example
Suppose f(t) is the function graphed below:
..
y
.
g
.
. . . . . . .
. x
.
2
. 4
. 6
.. 8 1f
. . 0.
∫ x
Let g(x) = f(t) dt. What can you say about g?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
28. Envisioning the area function
Example
Suppose f(t) is the function graphed below:
..
y
.
g
.
. . . . . . .
. x
.
2
. 4
. 6
.. 8 1f
. . 0.
∫ x
Let g(x) = f(t) dt. What can you say about g?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
29. Envisioning the area function
Example
Suppose f(t) is the function graphed below:
..
y
.
g
.
. . . . . . .
. x
.
2
. 4
. 6
.. 8 1f
. . 0.
∫ x
Let g(x) = f(t) dt. What can you say about g?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
30. Envisioning the area function
Example
Suppose f(t) is the function graphed below:
..
y
.
g
.
. . . . . . .
. x
.
2
. 4
. 6
.. 8 1f
. . 0.
∫ x
Let g(x) = f(t) dt. What can you say about g?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
31. Envisioning the area function
Example
Suppose f(t) is the function graphed below:
..
y
.
g
.
. . . . . . .
. x
.
2
. 4
. 6
.. 8 1f
. . 0.
∫ x
Let g(x) = f(t) dt. What can you say about g?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
32. features of g from f
..
y
Interval sign monotonicity monotonicity concavity
. of f of g of f of g
g
.
. . . . . . . [0, 2] + ↗ ↗ ⌣
. ..fx
. . .. . . 0
2 4 6 81 [2, 4.5] + ↗ ↘ ⌢
[4.5, 6] − ↘ ↘ ⌢
[6, 8] − ↘ ↗ ⌣
[8, 10] − ↘ → none
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 15 / 34
33. features of g from f
..
y
Interval sign monotonicity monotonicity concavity
. of f of g of f of g
g
.
. . . . . . . [0, 2] + ↗ ↗ ⌣
. ..fx
. . .. . . 0
2 4 6 81 [2, 4.5] + ↗ ↘ ⌢
[4.5, 6] − ↘ ↘ ⌢
[6, 8] − ↘ ↗ ⌣
[8, 10] − ↘ → none
We see that g is behaving a lot like an antiderivative of f.
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 15 / 34
34. 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).
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 16 / 34
35. Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
g(x + h) − g(x)
=
h
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 17 / 34
36. 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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 17 / 34
37. 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
So
g(x + h) − g(x)
mh ≤ ≤ Mh .
h
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 17 / 34
38. 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
So
g(x + h) − g(x)
mh ≤ ≤ Mh .
h
As h → 0, both mh and Mh tend to f(x).
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 17 / 34
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
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).
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 17 / 34
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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 18 / 34
41. Meet the Mathematician: Isaac Barrow
English, 1630-1677
Professor of Greek,
theology, and mathematics
at Cambridge
Had a famous student
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 19 / 34
42. Meet the Mathematician: Isaac Newton
English, 1643–1727
Professor at Cambridge
(England)
Philosophiae Naturalis
Principia Mathematica
published 1687
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 20 / 34
43. Meet the Mathematician: Gottfried Leibniz
German, 1646–1716
Eminent philosopher as
well as mathematician
Contemporarily disgraced
by the calculus priority
dispute
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 21 / 34
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.
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 22 / 34
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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 23 / 34
46. Differentiation of area functions
Example
∫ 3x
Let h(x) = t3 dt. What is h′ (x)?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 24 / 34
47. 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 4
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 24 / 34
48. 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 4
0
Solution (Using 1FTC)
∫ u
We can think of h as the composition g k, where g(u) = ◦ t3 dt and
0
k(x) = 3x.
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 24 / 34
49. 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 4
0
Solution (Using 1FTC)
∫ u
We can think of h as the composition g k, where g(u) = ◦ t3 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 = 81x3 .
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 24 / 34
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)
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 25 / 34
51. Example
∫ sin2 x
Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)?
0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 26 / 34
52. 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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 26 / 34
53. Example
∫ sin2 x
Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)?
3
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 27 / 34
54. Example
∫ sin2 x
Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)?
3
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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 27 / 34
55. Question
Why is
∫ sin2 x ∫ sin2 x
d d
(17t2 + 4t − 4) dt = (17t2 + 4t − 4) dt?
dx 0 dx 3
Or, why doesn’t the lower limit appear in the derivative?
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 28 / 34
56. Question
Why is
∫ sin2 x ∫ sin2 x
d d
(17t2 + 4t − 4) dt = (17t2 + 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
(17t + 4t − 4) dt = 2
(17t + 4t − 4) dt + (17t2 + 4t − 4) dt
0 0 3
So the two functions differ by a constant.
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 28 / 34
57. Example
∫ ex
Find the derivative of F(x) = sin4 t dt.
x3
. . . . . .
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58. 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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 29 / 34
59. 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 .
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 29 / 34
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.
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 30 / 34
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.
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 30 / 34
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.
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 30 / 34
63. Erf
Here’s a function with a funny name but an important role:
∫ x
2
e−t dt.
2
erf(x) = √
π 0
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 31 / 34
64. Erf
Here’s a function with a funny name but an important role:
∫ x
2
e−t dt.
2
erf(x) = √
π 0
It turns out erf is the shape of the bell curve.
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 31 / 34
65. Erf
Here’s a function with a funny name but an important role:
∫ x
2
e−t dt.
2
erf(x) = √
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) =
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 31 / 34
66. Erf
Here’s a function with a funny name but an important role:
∫ x
2
e−t dt.
2
erf(x) = √
π 0
It turns out erf is the shape of the bell curve. We can’t find erf(x),
2
explicitly, but we do know its derivative: erf′ (x) = √ e−x .
2
π
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 31 / 34
67. Erf
Here’s a function with a funny name but an important role:
∫ x
2
e−t dt.
2
erf(x) = √
π 0
It turns out erf is the shape of the bell curve. We can’t find erf(x),
2
explicitly, but we do know its derivative: erf′ (x) = √ e−x .
2
π
Example
d
Find erf(x2 ).
dx
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 31 / 34
68. Erf
Here’s a function with a funny name but an important role:
∫ x
2
e−t dt.
2
erf(x) = √
π 0
It turns out erf is the shape of the bell curve. We can’t find erf(x),
2
explicitly, but we do know its derivative: erf′ (x) = √ e−x .
2
π
Example
d
Find erf(x2 ).
dx
Solution
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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 31 / 34
69. 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.
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 32 / 34
70. Surplus by picture
c
. onsumer surplus
p
. rice (p)
s
. upply
.∗ .
p . . quilibrium
e
d
. emand f(q)
. .
.∗
q q
. uantity (q)
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 33 / 34
71. 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
. . . . . .
V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 34 / 34