. . SecƟon 5.4 The Fundamental Theorem of Calculus V63.0121.001: Calculus I Professor MaƩhew Leingang New York University May 2, 2011
Lesson 26: The Fundamental Theorem of Calculus (slides)
Dec 14, 2014
The fundamental theorem shows that differentiation and integration are inverse processes.
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Sec on 5.4The Fundamental Theorem of
Calculus
V63.0121.001: Calculus IProfessor Ma hew Leingang
New York University
May 2, 2011
Announcements
I Today: 5.4I Wednesday 5/4: 5.5I Monday 5/9: Review andMovie Day!
I Thursday 5/12: FinalExam, 2:00–3:50pm
ObjectivesI State and explain theFundamental Theorems ofCalculus
I Use the first fundamentaltheorem of calculus to findderiva ves of func ons definedas integrals.
I Compute the average value ofan integrable func on over aclosed interval.
OutlineRecall: The Evalua on Theorem a/k/a 2nd FTC
The First Fundamental Theorem of CalculusArea as a Func onStatement and proof of 1FTCBiographies
Differen a on of func ons defined by integrals“Contrived” examplesErfOther applica ons
The definite integral as a limitDefini onIf f is a func on defined on [a, b], the definite integral of f from a tob is the number ∫ b
af(x) dx = lim
n→∞
n∑i=1
f(ci)∆x
where∆x =b− an
, and for each i, xi = a+ i∆x, and ci is a point in[xi−1, xi].
The definite integral as a limit
TheoremIf f is con nuous on [a, b] or if f has only finitely many jumpdiscon nui es, then f is integrable on [a, b]; that is, the definite
integral∫ b
af(x) dx exists and is the same for any choice of ci.
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
af(x) dx = F(b)− F(a).
The Integral as Net Change
The Integral as Net Change
Corollary
If v(t) represents the velocity of a par cle moving rec linearly, then∫ t1
t0v(t) dt = s(t1)− s(t0).
The Integral as Net Change
Corollary
If MC(x) represents the marginal cost of making x units of a product,then
C(x) = C(0) +∫ x
0MC(q) dq.
The Integral as Net Change
Corollary
If ρ(x) represents the density of a thin rod at a distance of x from itsend, then the mass of the rod up to x is
m(x) =∫ x
0ρ(s) ds.
My first table of integrals..∫
[f(x) + g(x)] dx =∫
f(x) dx+∫
g(x) dx∫xn dx =
xn+1
n+ 1+ C (n ̸= −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∫
1xdx = 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
OutlineRecall: The Evalua on Theorem a/k/a 2nd FTC
The First Fundamental Theorem of CalculusArea as a Func onStatement and proof of 1FTCBiographies
Differen a on of func ons defined by integrals“Contrived” examplesErfOther applica ons
Area as a FunctionExample
Let f(t) = t3 and define g(x) =∫ x
0f(t) dt. Find g(x) and g′(x).
Solu on
..0.
x
Rn
Area as a FunctionExample
Let f(t) = t3 and define g(x) =∫ x
0f(t) dt. Find g(x) and g′(x).
Solu on
..0.
x
Dividing the interval [0, x] into n pieces gives∆t =xnand
ti = 0+ i∆t =ixn.
Rn
Area as a FunctionExample
Let f(t) = t3 and define g(x) =∫ x
0f(t) dt. Find g(x) and g′(x).
Solu on
..0.
x
Rn =xn· x
3
n3+
xn· (2x)
3
n3+ · · ·+ x
n· (nx)
3
n3
Area as a FunctionExample
Let f(t) = t3 and define g(x) =∫ x
0f(t) dt. Find g(x) and g′(x).
Solu on
..0.
x
Rn =xn· x
3
n3+
xn· (2x)
3
n3+ · · ·+ x
n· (nx)
3
n3
=x4
n4(13 + 23 + 33 + · · ·+ n3
)
Area as a FunctionExample
Let f(t) = t3 and define g(x) =∫ x
0f(t) dt. Find g(x) and g′(x).
Solu on
..0.
x
Rn =xn· x
3
n3+
xn· (2x)
3
n3+ · · ·+ x
n· (nx)
3
n3
=x4
n4(13 + 23 + 33 + · · ·+ n3
)=
x4
n4[12n(n+ 1)
]2
Area as a FunctionExample
Let f(t) = t3 and define g(x) =∫ x
0f(t) dt. Find g(x) and g′(x).
Solu on
..0.
x
Rn =x4n2(n+ 1)2
4n4
Area as a FunctionExample
Let f(t) = t3 and define g(x) =∫ x
0f(t) dt. Find g(x) and g′(x).
Solu on
..0.
x
Rn =x4n2(n+ 1)2
4n4
So g(x) = limx→∞
Rn =x4
4
and g′(x) = x3.
Area as a FunctionExample
Let f(t) = t3 and define g(x) =∫ x
0f(t) dt. Find g(x) and g′(x).
Solu on
..0.
x
Rn =x4n2(n+ 1)2
4n4
So g(x) = limx→∞
Rn =x4
4and g′(x) = x3.
The area function in generalLet f be a func on which is integrable (i.e., con nuous or withfinitely many jump discon nui es) 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 underthe curve.
I Ques on: What does f tell you about g?
Envisioning the area function
Example
Suppose f(t) is the func ongraphed to the right. Let
g(x) =∫ x
0f(t) dt
What can you say about g?...
x..
y
....f
..2
..4
..6
..8
..10
Envisioning the area function
Example
Suppose f(t) is the func ongraphed to the right. Let
g(x) =∫ x
0f(t) dt
What can you say about g?...
x..
y
....g
....f
..2
..4
..6
..8
..10
Envisioning the area function
Example
Suppose f(t) is the func ongraphed to the right. Let
g(x) =∫ x
0f(t) dt
What can you say about g?...
x..
y
....g
....f
..2
..4
..6
..8
..10
Envisioning the area function
Example
Suppose f(t) is the func ongraphed to the right. Let
g(x) =∫ x
0f(t) dt
What can you say about g?...
x..
y
....g
....f
..2
..4
..6
..8
..10
Envisioning the area function
Example
Suppose f(t) is the func ongraphed to the right. Let
g(x) =∫ x
0f(t) dt
What can you say about g?...
x..
y
....g
....f
..2
..4
..6
..8
..10
Envisioning the area function
Example
Suppose f(t) is the func ongraphed to the right. Let
g(x) =∫ x
0f(t) dt
What can you say about g?...
x..
y
....g
....f
..2
..4
..6
..8
..10
Envisioning the area function
Example
Suppose f(t) is the func ongraphed to the right. Let
g(x) =∫ x
0f(t) dt
What can you say about g?...
x..
y
....g
....f
..2
..4
..6
..8
..10
Envisioning the area function
Example
Suppose f(t) is the func ongraphed to the right. Let
g(x) =∫ x
0f(t) dt
What can you say about g?...
x..
y
....g
....f
..2
..4
..6
..8
..10
Envisioning the area function
Example
Suppose f(t) is the func ongraphed to the right. Let
g(x) =∫ x
0f(t) dt
What can you say about g?...
x..
y
....g
....f
..2
..4
..6
..8
..10
Envisioning the area function
Example
Suppose f(t) is the func ongraphed to the right. Let
g(x) =∫ x
0f(t) dt
What can you say about g?...
x..
y
....g
....f
..2
..4
..6
..8
..10
Envisioning the area function
Example
Suppose f(t) is the func ongraphed to the right. Let
g(x) =∫ x
0f(t) dt
What can you say about g?...
x..
y
....g
....f
..2
..4
..6
..8
..10
Envisioning the area function
Example
Suppose f(t) is the func ongraphed to the right. Let
g(x) =∫ x
0f(t) dt
What can you say about g?...
x..
y
....g
....f
..2
..4
..6
..8
..10
Features of g from f
...x
..
y
.... g....f
..2
..4
..6
..8
..10
Interval sign monotonicity monotonicity concavityof 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 an deriva ve of f.
Features of g from f
...x
..
y
.... g....f
..2
..4
..6
..8
..10
Interval sign monotonicity monotonicity concavityof 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 an deriva ve of f.
Another Big Time TheoremTheorem (The First Fundamental Theorem of Calculus)
Let f be an integrable func on on [a, b] and define
g(x) =∫ x
af(t) dt.
If f is con nuous at x in (a, b), then g is differen able at x and
g′(x) = f(x).
Proving the Fundamental TheoremProof.
Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=
1h
∫ x+h
xf(t) dt.
LetMh be the maximum value of f on [x, x+ h], and letmh theminimum value of f on [x, x+ h]. From §5.2 we have
mh · h ≤
∫ x+h
xf(t) dt
≤ Mh · h
=⇒ mh ≤g(x+ h)− g(x)
h≤ Mh.
As h → 0, bothmh andMh tend to f(x).
Proving the Fundamental TheoremProof.
Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=1h
∫ x+h
xf(t) dt.
LetMh be the maximum value of f on [x, x+ h], and letmh theminimum value of f on [x, x+ h]. From §5.2 we have
mh · h ≤
∫ x+h
xf(t) dt
≤ Mh · h
=⇒ mh ≤g(x+ h)− g(x)
h≤ Mh.
As h → 0, bothmh andMh tend to f(x).
Proving the Fundamental TheoremProof.
Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=1h
∫ x+h
xf(t) dt.
LetMh be the maximum value of f on [x, x+ h], and letmh theminimum value of f on [x, x+ h].
From §5.2 we have
mh · h ≤
∫ x+h
xf(t) dt
≤ Mh · h
=⇒ mh ≤g(x+ h)− g(x)
h≤ Mh.
As h → 0, bothmh andMh tend to f(x).
Proving the Fundamental TheoremProof.
Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=1h
∫ x+h
xf(t) dt.
LetMh be the maximum value of f on [x, x+ h], and letmh theminimum value of f on [x, x+ h]. From §5.2 we have
mh · h ≤
∫ x+h
xf(t) dt
≤ Mh · h
=⇒ mh ≤g(x+ h)− g(x)
h≤ Mh.
As h → 0, bothmh andMh tend to f(x).
Proving the Fundamental TheoremProof.
Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=1h
∫ x+h
xf(t) dt.
LetMh be the maximum value of f on [x, x+ h], and letmh theminimum value of f on [x, x+ h]. From §5.2 we have
mh · h ≤
∫ x+h
xf(t) dt ≤ Mh · h
=⇒ mh ≤g(x+ h)− g(x)
h≤ Mh.
As h → 0, bothmh andMh tend to f(x).
Proving the Fundamental TheoremProof.
Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=1h
∫ x+h
xf(t) dt.
LetMh be the maximum value of f on [x, x+ h], and letmh theminimum value of f on [x, x+ h]. From §5.2 we have
mh · h ≤∫ x+h
xf(t) dt ≤ Mh · h
=⇒ mh ≤g(x+ h)− g(x)
h≤ Mh.
As h → 0, bothmh andMh tend to f(x).
Proving the Fundamental TheoremProof.
From §5.2 we have
mh · h ≤∫ x+h
xf(t) dt ≤ Mh · h
=⇒ mh ≤g(x+ h)− g(x)
h≤ Mh.
As h → 0, bothmh andMh tend to f(x).
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
About Mh and mh
I Since f is con nuous at x,and x+ h → x, we have
limh→0
Mh = f(x)
limh→0
mh = f(x)
I This is not necessarilytrue when f is notcon nuous.
.
..
f(x)
..x
...
f(x)
..x
..
Mh
...x+ h
.
mh
.
Meet the MathematicianJames Gregory
I Sco sh, 1638-1675I Astronomer and GeometerI Conceived transcendentalnumbers and found evidencethat π was transcendental
I Proved a geometric version of1FTC as a lemma but didn’t takeit further
Meet the MathematicianIsaac Barrow
I English, 1630-1677I Professor of Greek, theology,and mathema cs at Cambridge
I Had a famous student
Meet the MathematicianIsaac Newton
I English, 1643–1727I Professor at Cambridge(England)
I invented calculus 1665–66I Tractus de QuadraturaCurvararum published 1704
Meet the MathematicianGottfried Leibniz
I German, 1646–1716I Eminent philosopher as well asmathema cian
I invented calculus 1672–1676I published in papers 1684 and1686
Differentiation and Integration asreverse processes
Pu ng together 1FTC and 2FTC, we get a beau ful rela onshipbetween the two fundamental concepts in calculus.Theorem (The Fundamental Theorem(s) of Calculus)
I. If f is a con nuous func on, then
ddx
∫ x
af(t) dt = f(x)
So the deriva ve of the integral is the original func on.
Differentiation and Integration asreverse processes
Pu ng together 1FTC and 2FTC, we get a beau ful rela onshipbetween the two fundamental concepts in calculus.Theorem (The Fundamental Theorem(s) of Calculus)
II. If f is a differen able func on, then∫ b
af′(x) dx = f(b)− f(a).
So the integral of the deriva ve of is (an evalua on of) theoriginal func on.
OutlineRecall: The Evalua on Theorem a/k/a 2nd FTC
The First Fundamental Theorem of CalculusArea as a Func onStatement and proof of 1FTCBiographies
Differen a on of func ons defined by integrals“Contrived” examplesErfOther applica ons
Differentiation of area functionsExample
Let h(x) =∫ 3x
0t3 dt. What is h′(x)?
Solu on (Using 1FTC)
We can think of h as the composi on 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.
Differentiation of area functionsExample
Let h(x) =∫ 3x
0t3 dt. What is h′(x)?
Solu on (Using 2FTC)
h(x) =t4
4
∣∣∣∣3x0=
14(3x)4 = 1
4 · 81x4, so h′(x) = 81x3.
Solu on (Using 1FTC)
We can think of h as the composi on 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.
Differentiation of area functionsExample
Let h(x) =∫ 3x
0t3 dt. What is h′(x)?
Solu on (Using 1FTC)
We can think of h as the composi on 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.
Differentiation of area functionsExample
Let h(x) =∫ 3x
0t3 dt. What is h′(x)?
Solu on (Using 1FTC)
We can think of h as the composi on 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.
Differentiation of area functions, in generalI by 1FTC
ddx
∫ k(x)
af(t) dt = f(k(x))k′(x)
I by reversing the order of integra on:
ddx
∫ b
h(x)f(t) dt = − d
dx
∫ h(x)
bf(t) dt = −f(h(x))h′(x)
I by combining the two above:
ddx
∫ k(x)
h(x)f(t) dt =
ddx
(∫ k(x)
0f(t) dt+
∫ 0
h(x)f(t) dt
)= f(k(x))k′(x)− f(h(x))h′(x)
Another ExampleExample
Let h(x) =∫ sin2 x
0(17t2 + 4t− 4) dt. What is h′(x)?
Another ExampleExample
Let h(x) =∫ sin2 x
0(17t2 + 4t− 4) dt. What is h′(x)?
Solu on (2FTC)
h(x) =173t3 + 2t2 − 4t
∣∣∣∣sin2 x3
=173(sin2 x)3 + 2(sin2 x)2 − 4(sin2 x)
=173
sin6 x+ 2 sin4 x− 4 sin2 x
h′(x) =(17 · 63
sin5 x+ 2 · 4 sin3 x− 4 · 2 sin x)cos x
Another ExampleExample
Let h(x) =∫ sin2 x
0(17t2 + 4t− 4) dt. What is h′(x)?
Solu on (1FTC)
ddx
∫ sin2 x
0(17t2 + 4t− 4) dt =
(17(sin2 x)2 + 4(sin2 x)− 4
)· ddx
sin2 x
=(17 sin4 x+ 4 sin2 x− 4
)· 2 sin x cos x
A Similar ExampleExample
Let h(x) =∫ sin2 x
3(17t2 + 4t− 4) dt. What is h′(x)?
Solu onWe have
ddx
∫ sin2 x
3(17t2 + 4t− 4) dt =
(17(sin2 x)2 + 4(sin2 x)− 4
)· ddx
sin2 x
=(17 sin4 x+ 4 sin2 x− 4
)· 2 sin x cos x
A Similar ExampleExample
Let h(x) =∫ sin2 x
3(17t2 + 4t− 4) dt. What is h′(x)?
Solu onWe have
ddx
∫ sin2 x
3(17t2 + 4t− 4) dt =
(17(sin2 x)2 + 4(sin2 x)− 4
)· ddx
sin2 x
=(17 sin4 x+ 4 sin2 x− 4
)· 2 sin x cos x
CompareQues on
Why isddx
∫ sin2 x
0(17t2 + 4t− 4) dt =
ddx
∫ sin2 x
3(17t2 + 4t− 4) dt?
Or, why doesn’t the lower limit appear in the deriva ve?
Answer∫ sin2 x
0(17t2+4t−4) dt =
∫ 3
0(17t2+4t−4) dt+
∫ sin2 x
3(17t2+4t−4) dt
So the two func ons differ by a constant.
CompareQues on
Why isddx
∫ sin2 x
0(17t2 + 4t− 4) dt =
ddx
∫ sin2 x
3(17t2 + 4t− 4) dt?
Or, why doesn’t the lower limit appear in the deriva ve?
Answer∫ sin2 x
0(17t2+4t−4) dt =
∫ 3
0(17t2+4t−4) dt+
∫ sin2 x
3(17t2+4t−4) dt
So the two func ons differ by a constant.
The Full NastyExample
Find the deriva ve of F(x) =∫ ex
x3sin4 t dt.
Solu on
ddx
∫ ex
x3sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2
No ce here it’s much easier than finding an an deriva ve for sin4.
The Full NastyExample
Find the deriva ve of F(x) =∫ ex
x3sin4 t dt.
Solu on
ddx
∫ ex
x3sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2
No ce here it’s much easier than finding an an deriva ve for sin4.
The Full NastyExample
Find the deriva ve of F(x) =∫ ex
x3sin4 t dt.
Solu on
ddx
∫ ex
x3sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2
No ce here it’s much easier than finding an an deriva ve for sin4.
Why use 1FTC?Ques on
Why would we use 1FTC to find the deriva ve of an integral? Itseems like confusion for its own sake.
Answer
I Some func ons are difficult or impossible to integrate inelementary terms.
I Some func ons are naturally defined in terms of otherintegrals.
Why use 1FTC?Ques on
Why would we use 1FTC to find the deriva ve of an integral? Itseems like confusion for its own sake.
Answer
I Some func ons are difficult or impossible to integrate inelementary terms.
I Some func ons are naturally defined in terms of otherintegrals.
Why use 1FTC?Ques on
Why would we use 1FTC to find the deriva ve of an integral? Itseems like confusion for its own sake.
Answer
I Some func ons are difficult or impossible to integrate inelementary terms.
I Some func ons are naturally defined in terms of otherintegrals.
Erferf(x) =
2√π
∫ x
0e−t2 dt
I erf measures area the bell curve.I We can’t find erf(x), explicitly,but we do know its deriva ve:
erf′(x) =
2√πe−x2.
...x
.
erf(x)
Erferf(x) =
2√π
∫ x
0e−t2 dt
I erf measures area the bell curve.
I We can’t find erf(x), explicitly,but we do know its deriva ve:
erf′(x) =
2√πe−x2.
...x
.
erf(x)
Erferf(x) =
2√π
∫ x
0e−t2 dt
I erf measures area the bell curve.I We can’t find erf(x), explicitly,but we do know its deriva ve:
erf′(x) =
2√πe−x2.
...x
.
erf(x)
Erferf(x) =
2√π
∫ x
0e−t2 dt
I erf measures area the bell curve.I We can’t find erf(x), explicitly,but we do know its deriva ve:
erf′(x) =2√πe−x2.
...x
.
erf(x)
Example of erfExample
Findddx
erf(x2).
Solu onBy the chain rule we have
ddx
erf(x2) = erf′(x2)ddx
x2 =2√πe−(x2)22x =
4√πxe−x4.
Example of erfExample
Findddx
erf(x2).
Solu onBy the chain rule we have
ddx
erf(x2) = erf′(x2)ddx
x2 =2√πe−(x2)22x =
4√πxe−x4.
Other functions defined by integralsI The future value of an asset:
FV(t) =∫ ∞
tπ(s)e−rs ds
where π(s) is the profitability at me s and r is the discountrate.
I The consumer surplus of a good:
CS(q∗) =∫ q∗
0(f(q)− p∗) dq
where f(q) is the demand func on and p∗ and q∗ theequilibrium price and quan ty.
Surplus by picture
..quan ty (q)
.
price (p)
Surplus by picture
..quan ty (q)
.
price (p)
..
demand f(q)
Surplus by picture
..quan ty (q)
.
price (p)
..
demand f(q).
supply
Surplus by picture
..quan ty (q)
.
price (p)
..
demand f(q).
supply
.
equilibrium
..q∗
..
p∗
Surplus by picture
..quan ty (q)
.
price (p)
..
demand f(q).
market revenue
.
supply
.
equilibrium
..q∗
..
p∗
Surplus by picture
..quan ty (q)
.
price (p)
..
demand f(q).
market revenue
.
supply
.
equilibrium
..q∗
..
p∗
.
consumer surplus
Surplus by picture
..quan ty (q)
.
price (p)
..
demand f(q).
supply
.
equilibrium
..q∗
..
p∗
.
consumer surplus
.
producer surplus
SummaryI Func ons defined as integrals can be differen ated using thefirst FTC:
ddx
∫ x
af(t) dt = f(x)
I The two FTCs link the two major processes in calculus:differen a on and integra on∫
F′(x) dx = F(x) + C
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