Section 5.4The Fundamental Theorem of Calculus
V63.0121.002.2010Su, Calculus I
New York University
June 21, 2010
Announcements
I
Announcements
I
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 2 / 33
Objectives
I State and explain theFundemental Theorems ofCalculus
I Use the first fundamentaltheorem of calculus to findderivatives of functionsdefined as integrals.
I Compute the average valueof an integrable functionover a closed interval.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 3 / 33
Outline
Recall: The Evaluation Theorem a/k/a 2FTC
The First Fundamental Theorem of CalculusThe Area FunctionStatement and proof of 1FTCBiographies
Differentiation of functions defined by integrals“Contrived” examplesErfOther applications
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 4 / 33
The definite integral as a limit
Definition
If f is a function defined on [a, b], the definite integral of f from a to bis the number ∫ b
af (x) dx = lim
∆x→0
n∑i=1
f (ci ) ∆x
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 5 / 33
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
af (x) dx = F (b)− F (a).
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 6 / 33
The Integral as Total Change
Another way to state this theorem is:∫ b
aF ′(x) dx = F (b)− F (a),
or the integral of a derivative along an interval is the total change betweenthe 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
The Integral as Total Change
Another way to state this theorem is:∫ b
aF ′(x) dx = F (b)− F (a),
or the integral of a derivative along an interval is the total change betweenthe sides of that interval. This has many ramifications:
Theorem
If v(t) represents the velocity of a particle moving rectilinearly, then∫ t1
t0
v(t) dt = s(t1)− s(t0).
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 7 / 33
The Integral as Total Change
Another way to state this theorem is:∫ b
aF ′(x) dx = F (b)− F (a),
or the integral of a derivative along an interval is the total change betweenthe sides of that interval. This has many ramifications:
Theorem
If MC (x) represents the marginal cost of making x units of a product, then
C (x) = C (0) +
∫ x
0MC (q) dq.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 7 / 33
The Integral as Total Change
Another way to state this theorem is:∫ b
aF ′(x) dx = F (b)− F (a),
or the integral of a derivative along an interval is the total change betweenthe 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
m(x) =
∫ x
0ρ(s) ds.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 7 / 33
My first table of integrals
∫[f (x) + g(x)] dx =
∫f (x) dx +
∫g(x) dx∫
xn dx =xn+1
n + 1+ C (n 6= −1)∫
ex dx = ex + C∫sin x dx = − cos x + C∫cos x dx = sin x + C∫
sec2 x dx = tan x + C∫sec x tan x dx = sec x + C∫
1
1 + x2dx = arctan x + C
∫cf (x) dx = c
∫f (x) dx∫
1
xdx = ln |x |+ C∫
ax dx =ax
ln a+ C∫
csc2 x dx = − cot x + C∫csc x cot x dx = − csc x + C∫
1√1− x2
dx = arcsin x + C
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 8 / 33
Outline
Recall: The Evaluation Theorem a/k/a 2FTC
The First Fundamental Theorem of CalculusThe Area FunctionStatement and proof of 1FTCBiographies
Differentiation of functions defined by integrals“Contrived” examplesErfOther applications
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 9 / 33
An area function
Let f (t) = t3 and define g(x) =
∫ x
0f (t) dt. Can we evaluate the integral
in g(x)?
0 x
Dividing the interval [0, x ] into n pieces
gives ∆t =x
nand ti = 0 + i∆t =
ix
n. So
Rn =x
n· x3
n3+
x
n· (2x)3
n3+ · · ·+ x
n· (nx)3
n3
=x4
n4
(13 + 23 + 33 + · · ·+ n3
)=
x4
n4
[12 n(n + 1)
]2=
x4n2(n + 1)2
4n4→ x4
4
as n→∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 10 / 33
An area function
Let f (t) = t3 and define g(x) =
∫ x
0f (t) dt. Can we evaluate the integral
in g(x)?
0 x
Dividing the interval [0, x ] into n pieces
gives ∆t =x
nand ti = 0 + i∆t =
ix
n. So
Rn =x
n· x3
n3+
x
n· (2x)3
n3+ · · ·+ x
n· (nx)3
n3
=x4
n4
(13 + 23 + 33 + · · ·+ n3
)=
x4
n4
[12 n(n + 1)
]2=
x4n2(n + 1)2
4n4→ x4
4
as n→∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 10 / 33
An area function, continued
So
g(x) =x4
4.
This means thatg ′(x) = x3.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 11 / 33
An area function, continued
So
g(x) =x4
4.
This means thatg ′(x) = x3.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 11 / 33
The area function
Let f be a function which is integrable (i.e., continuous or with finitelymany jump discontinuities) on [a, b]. Define
g(x) =
∫ x
af (t) dt.
I The variable is x ; t is a “dummy” variable that’s integrated over.
I Picture changing x and taking more of less of the region under thecurve.
I 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
Envisioning the area function
Example
Suppose f (t) is the function graphed below:
x
y
f2 4 6 8 10
Let g(x) =
∫ x
0f (t) dt. What can you say about g?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
Envisioning the area function
Example
Suppose f (t) is the function graphed below:
x
y
g
f2 4 6 8 10
Let g(x) =
∫ x
0f (t) dt. What can you say about g?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
Envisioning the area function
Example
Suppose f (t) is the function graphed below:
x
y
g
f2 4 6 8 10
Let g(x) =
∫ x
0f (t) dt. What can you say about g?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
Envisioning the area function
Example
Suppose f (t) is the function graphed below:
x
y
g
f2 4 6 8 10
Let g(x) =
∫ x
0f (t) dt. What can you say about g?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
Envisioning the area function
Example
Suppose f (t) is the function graphed below:
x
y
g
f2 4 6 8 10
Let g(x) =
∫ x
0f (t) dt. What can you say about g?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
Envisioning the area function
Example
Suppose f (t) is the function graphed below:
x
y
g
f2 4 6 8 10
Let g(x) =
∫ x
0f (t) dt. What can you say about g?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
Envisioning the area function
Example
Suppose f (t) is the function graphed below:
x
y
g
f2 4 6 8 10
Let g(x) =
∫ x
0f (t) dt. What can you say about g?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
Envisioning the area function
Example
Suppose f (t) is the function graphed below:
x
y
g
f2 4 6 8 10
Let g(x) =
∫ x
0f (t) dt. What can you say about g?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
Envisioning the area function
Example
Suppose f (t) is the function graphed below:
x
y
g
f2 4 6 8 10
Let g(x) =
∫ x
0f (t) dt. What can you say about g?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
Envisioning the area function
Example
Suppose f (t) is the function graphed below:
x
y
g
f2 4 6 8 10
Let g(x) =
∫ x
0f (t) dt. What can you say about g?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
Envisioning the area function
Example
Suppose f (t) is the function graphed below:
x
y
g
f2 4 6 8 10
Let g(x) =
∫ x
0f (t) dt. What can you say about g?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
Envisioning the area function
Example
Suppose f (t) is the function graphed below:
x
y
g
f2 4 6 8 10
Let g(x) =
∫ x
0f (t) dt. What can you say about g?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
features of g from f
x
y
g
f2 4 6 8 10
Interval sign monotonicity monotonicity concavity
of f of g of f of g
[0, 2] + ↗ ↗ ^
[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
features of g from f
x
y
g
f2 4 6 8 10
Interval sign monotonicity monotonicity concavity
of f of g of f of g
[0, 2] + ↗ ↗ ^
[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
Another Big Time Theorem
Theorem (The First Fundamental Theorem of Calculus)
Let f be an integrable function on [a, b] and define
g(x) =
∫ x
af (t) dt.
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
Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and let mh the minimumvalue of f on [x , x + h]. From §5.2 we have
mh · h ≤
∫ x+h
xf (t) dt
≤ Mh · h
So
mh ≤g(x + h)− g(x)
h≤ Mh.
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
Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and let mh the minimumvalue of f on [x , x + h]. From §5.2 we have
mh · h ≤
∫ x+h
xf (t) dt
≤ Mh · h
So
mh ≤g(x + h)− g(x)
h≤ Mh.
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
Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and let mh the minimumvalue of f on [x , x + h]. From §5.2 we have
mh · h ≤
∫ x+h
xf (t) dt
≤ Mh · h
So
mh ≤g(x + h)− g(x)
h≤ Mh.
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
Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and let mh the minimumvalue of f on [x , x + h]. From §5.2 we have
mh · h ≤
∫ x+h
xf (t) dt ≤ Mh · h
So
mh ≤g(x + h)− g(x)
h≤ Mh.
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
Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and let mh the minimumvalue of f on [x , x + h]. From §5.2 we have
mh · h ≤∫ x+h
xf (t) dt ≤ Mh · h
So
mh ≤g(x + h)− g(x)
h≤ Mh.
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
Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and let mh the minimumvalue of f on [x , x + h]. From §5.2 we have
mh · h ≤∫ x+h
xf (t) dt ≤ Mh · h
So
mh ≤g(x + h)− g(x)
h≤ Mh.
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
Proving the Fundamental Theorem
Proof.
Let h > 0 be given so that x + h < b. We have
g(x + h)− g(x)
h=
1
h
∫ x+h
xf (t) dt.
Let Mh be the maximum value of f on [x , x + h], and let mh the minimumvalue of f on [x , x + h]. From §5.2 we have
mh · h ≤∫ x+h
xf (t) dt ≤ Mh · h
So
mh ≤g(x + h)− g(x)
h≤ Mh.
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
Meet the Mathematician: James Gregory
I Scottish, 1638-1675
I Astronomer and Geometer
I Conceived transcendentalnumbers and found evidencethat π was transcendental
I Proved a geometric versionof 1FTC as a lemma butdidn’t take it further
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 17 / 33
Meet the Mathematician: Isaac Barrow
I English, 1630-1677
I Professor of Greek, theology,and mathematics atCambridge
I Had a famous student
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 18 / 33
Meet the Mathematician: Isaac Newton
I English, 1643–1727
I Professor at Cambridge(England)
I Philosophiae NaturalisPrincipia Mathematicapublished 1687
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 19 / 33
Meet the Mathematician: Gottfried Leibniz
I German, 1646–1716
I Eminent philosopher as wellas mathematician
I Contemporarily disgraced bythe calculus priority dispute
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 20 / 33
Differentiation and Integration as reverse processes
Putting together 1FTC and 2FTC, we get a beautiful relationship betweenthe two fundamental concepts in calculus.
Theorem (The Fundamental Theorem(s) of Calculus)
I. If f is a continuous function, then
d
dx
∫ x
af (t) dt = f (x)
So the derivative of the integral is the original function.
II. If f is a differentiable function, then∫ b
af ′(x) dx = f (b)− f (a).
So the integral of the derivative of is (an evaluation of) the originalfunction.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 21 / 33
Outline
Recall: The Evaluation Theorem a/k/a 2FTC
The First Fundamental Theorem of CalculusThe Area FunctionStatement and proof of 1FTCBiographies
Differentiation of functions defined by integrals“Contrived” examplesErfOther applications
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 22 / 33
Differentiation of area functions
Example
Let h(x) =
∫ 3x
0t3 dt. What is h′(x)?
Solution (Using 2FTC)
h(x) =t4
4
∣∣∣∣3x0
=1
4(3x)4 = 1
4 · 81x4, so h′(x) = 81x3.
Solution (Using 1FTC)
We can think of h as the composition g ◦ k, where g(u) =
∫ u
0t3 dt 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 23 / 33
Differentiation of area functions
Example
Let h(x) =
∫ 3x
0t3 dt. What is h′(x)?
Solution (Using 2FTC)
h(x) =t4
4
∣∣∣∣3x0
=1
4(3x)4 = 1
4 · 81x4, so h′(x) = 81x3.
Solution (Using 1FTC)
We can think of h as the composition g ◦ k, where g(u) =
∫ u
0t3 dt 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 23 / 33
Differentiation of area functions
Example
Let h(x) =
∫ 3x
0t3 dt. What is h′(x)?
Solution (Using 2FTC)
h(x) =t4
4
∣∣∣∣3x0
=1
4(3x)4 = 1
4 · 81x4, so h′(x) = 81x3.
Solution (Using 1FTC)
We can think of h as the composition g ◦ k, where g(u) =
∫ u
0t3 dt 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 23 / 33
Differentiation of area functions
Example
Let h(x) =
∫ 3x
0t3 dt. What is h′(x)?
Solution (Using 2FTC)
h(x) =t4
4
∣∣∣∣3x0
=1
4(3x)4 = 1
4 · 81x4, so h′(x) = 81x3.
Solution (Using 1FTC)
We can think of h as the composition g ◦ k, where g(u) =
∫ u
0t3 dt 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 23 / 33
Differentiation of area functions, in general
I by 1FTCd
dx
∫ k(x)
af (t) dt = f (k(x))k ′(x)
I by reversing the order of integration:
d
dx
∫ b
h(x)f (t) dt = − d
dx
∫ h(x)
bf (t) dt = −f (h(x))h′(x)
I by combining the two above:
d
dx
∫ k(x)
h(x)f (t) dt =
d
dx
(∫ k(x)
0f (t) dt +
∫ 0
h(x)f (t) dt
)= 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
Another Example
Example
Let h(x) =
∫ sin2 x
0(17t2 + 4t − 4) dt. What is h′(x)?
Solution
We have
d
dx
∫ sin2 x
0(17t2 + 4t − 4) dt
=(17(sin2 x)2 + 4(sin2 x)− 4
)· d
dxsin2 x
=(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
Another Example
Example
Let h(x) =
∫ sin2 x
0(17t2 + 4t − 4) dt. What is h′(x)?
Solution
We have
d
dx
∫ sin2 x
0(17t2 + 4t − 4) dt
=(17(sin2 x)2 + 4(sin2 x)− 4
)· d
dxsin2 x
=(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
A Similar Example
Example
Let h(x) =
∫ sin2 x
3(17t2 + 4t − 4) dt. What is h′(x)?
Solution
We have
d
dx
∫ sin2 x
0(17t2 + 4t − 4) dt
=(17(sin2 x)2 + 4(sin2 x)− 4
)· d
dxsin2 x
=(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
A Similar Example
Example
Let h(x) =
∫ sin2 x
3(17t2 + 4t − 4) dt. What is h′(x)?
Solution
We have
d
dx
∫ sin2 x
0(17t2 + 4t − 4) dt
=(17(sin2 x)2 + 4(sin2 x)− 4
)· d
dxsin2 x
=(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
Compare
Question
Why is
d
dx
∫ sin2 x
0(17t2 + 4t − 4) dt =
d
dx
∫ sin2 x
3(17t2 + 4t − 4) dt?
Or, why doesn’t the lower limit appear in the derivative?
Answer
Because∫ sin2 x
0(17t2 + 4t−4) dt =
∫ 3
0(17t2 + 4t−4) dt +
∫ sin2 x
3(17t2 + 4t−4) dt
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
Compare
Question
Why is
d
dx
∫ sin2 x
0(17t2 + 4t − 4) dt =
d
dx
∫ sin2 x
3(17t2 + 4t − 4) dt?
Or, why doesn’t the lower limit appear in the derivative?
Answer
Because∫ sin2 x
0(17t2 + 4t−4) dt =
∫ 3
0(17t2 + 4t−4) dt +
∫ sin2 x
3(17t2 + 4t−4) dt
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
The Full Nasty
Example
Find the derivative of F (x) =
∫ ex
x3
sin4 t dt.
Solution
d
dx
∫ ex
x3
sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2
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
The Full Nasty
Example
Find the derivative of F (x) =
∫ ex
x3
sin4 t dt.
Solution
d
dx
∫ ex
x3
sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2
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
The Full Nasty
Example
Find the derivative of F (x) =
∫ ex
x3
sin4 t dt.
Solution
d
dx
∫ ex
x3
sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2
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
Why use 1FTC?
Question
Why would we use 1FTC to find the derivative of an integral? It seemslike confusion for its own sake.
Answer
I Some functions are difficult or impossible to integrate in elementaryterms.
I 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
Why use 1FTC?
Question
Why would we use 1FTC to find the derivative of an integral? It seemslike confusion for its own sake.
Answer
I Some functions are difficult or impossible to integrate in elementaryterms.
I 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
Why use 1FTC?
Question
Why would we use 1FTC to find the derivative of an integral? It seemslike confusion for its own sake.
Answer
I Some functions are difficult or impossible to integrate in elementaryterms.
I 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
Erf
Here’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t
2dt.
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) =2√π
e−x2.
Example
Findd
dxerf(x2).
Solution
By the chain rule we have
d
dxerf(x2) = erf ′(x2)
d
dxx2 =
2√π
e−(x2)22x =
4√π
xe−x4.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33
Erf
Here’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t
2dt.
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) =2√π
e−x2.
Example
Findd
dxerf(x2).
Solution
By the chain rule we have
d
dxerf(x2) = erf ′(x2)
d
dxx2 =
2√π
e−(x2)22x =
4√π
xe−x4.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33
Erf
Here’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t
2dt.
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) =
2√π
e−x2.
Example
Findd
dxerf(x2).
Solution
By the chain rule we have
d
dxerf(x2) = erf ′(x2)
d
dxx2 =
2√π
e−(x2)22x =
4√π
xe−x4.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33
Erf
Here’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t
2dt.
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) =2√π
e−x2.
Example
Findd
dxerf(x2).
Solution
By the chain rule we have
d
dxerf(x2) = erf ′(x2)
d
dxx2 =
2√π
e−(x2)22x =
4√π
xe−x4.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33
Erf
Here’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t
2dt.
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) =2√π
e−x2.
Example
Findd
dxerf(x2).
Solution
By the chain rule we have
d
dxerf(x2) = erf ′(x2)
d
dxx2 =
2√π
e−(x2)22x =
4√π
xe−x4.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33
Erf
Here’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t
2dt.
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) =2√π
e−x2.
Example
Findd
dxerf(x2).
Solution
By the chain rule we have
d
dxerf(x2) = erf ′(x2)
d
dxx2 =
2√π
e−(x2)22x =
4√π
xe−x4.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33
Other functions defined by integrals
I The future value of an asset:
FV (t) =
∫ ∞t
π(s)e−rs ds
where π(s) is the profitability at time s and r is the discount rate.
I The consumer surplus of a good:
CS(q∗) =
∫ q∗
0(f (q)− p∗) dq
where f (q) is the demand function and p∗ and q∗ the equilibriumprice and quantity.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 31 / 33
Surplus by picture
quantity (q)
price (p)
demand f (q)
market revenue
supply
equilibrium
q∗
p∗
consumer surplus
producer surplus
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
Surplus by picture
quantity (q)
price (p)
demand f (q)
market revenue
supply
equilibrium
q∗
p∗
consumer surplus
producer surplus
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
Surplus by picture
quantity (q)
price (p)
demand f (q)
market revenue
supply
equilibrium
q∗
p∗
consumer surplus
producer surplus
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
Surplus by picture
quantity (q)
price (p)
demand f (q)
market revenue
supply
equilibrium
q∗
p∗
consumer surplus
producer surplus
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
Surplus by picture
quantity (q)
price (p)
demand f (q)
market revenue
supply
equilibrium
q∗
p∗
consumer surplus
producer surplus
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
Surplus by picture
quantity (q)
price (p)
demand f (q)
market revenue
supply
equilibrium
q∗
p∗
consumer surplus
producer surplus
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
Surplus by picture
quantity (q)
price (p)
demand f (q)
market revenue
supply
equilibrium
q∗
p∗
consumer surplus
producer surplus
V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
Summary
I Functions defined as integrals can be differentiated using the firstFTC:
d
dx
∫ x
af (t) dt = f (x)
I The two FTCs link the two major processes in calculus: differentiationand integration ∫
F ′(x) dx = F (x) + C
I 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