..
Section 5.4The Fundamental Theorem of Calculus
V63.0121.021, Calculus I
New York University
December 9, 2010
Announcements
I Today: Section 5.4I ”Thursday,” December 14: Section 5.5I ”Monday,” December 15: (WWH 109, 12:30–1:45pm) Review and
Movie Day!I Monday, December 20, 12:00–1:50pm: Final Exam (location still
TBD). . . . . .
. . . . . .
Announcements
I Today: Section 5.4I ”Thursday,” December 14:
Section 5.5I ”Monday,” December 15:
(WWH 109,12:30–1:45pm) Reviewand Movie Day!
I Monday, December 20,12:00–1:50pm: Final Exam(location still TBD)
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 2 / 32
. . . . . .
Objectives
I State and explain theFundemental Theorems ofCalculus
I Use the first fundamentaltheorem of calculus to findderivatives of functionsdefined as integrals.
I Compute the averagevalue of an integrablefunction over a closedinterval.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 3 / 32
. . . . . .
Outline
Recall: The Evaluation Theorem a/k/a 2nd FTC
The First Fundamental Theorem of CalculusArea as a FunctionStatement and proof of 1FTCBiographies
Differentiation of functions defined by integrals“Contrived” examplesErfOther applications
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 4 / 32
. . . . . .
The definite integral as a limit
DefinitionIf 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
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 5 / 32
. . . . . .
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).
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 6 / 32
. . . . . .
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 changebetween the sides of that interval. This has many ramifications:
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 7 / 32
. . . . . .
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 changebetween the sides of that interval. This has many ramifications:
TheoremIf v(t) represents the velocity of a particle moving rectilinearly, then∫ t1
t0v(t)dt = s(t1)− s(t0).
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 7 / 32
. . . . . .
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 changebetween the sides of that interval. This has many ramifications:
TheoremIf MC(x) represents the marginal cost of making x units of a product,then
C(x) = C(0) +∫ x
0MC(q)dq.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 7 / 32
. . . . . .
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 changebetween the sides of that interval. This has many ramifications:
TheoremIf 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.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 7 / 32
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 8 / 32
. . . . . .
Outline
Recall: The Evaluation Theorem a/k/a 2nd FTC
The First Fundamental Theorem of CalculusArea as a FunctionStatement and proof of 1FTCBiographies
Differentiation of functions defined by integrals“Contrived” examplesErfOther applications
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 9 / 32
. . . . . .
Area as a Function
Example
Let f(t) = t3 and define g(x) =∫ x
0f(t)dt. Find g(x) and g′(x).
Solution
..0.
x
Dividing the interval [0, x] into n pieces
gives ∆t =xnand ti = 0+ i∆t =
ixn. So
Rn
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 10 / 32
. . . . . .
Area as a Function
Example
Let f(t) = t3 and define g(x) =∫ x
0f(t)dt. Find g(x) and g′(x).
Solution
..0.
x
Dividing the interval [0, x] into n pieces
gives ∆t =xnand ti = 0+ i∆t =
ixn.
So
Rn
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 10 / 32
. . . . . .
Area as a Function
Example
Let f(t) = t3 and define g(x) =∫ x
0f(t)dt. Find g(x) and g′(x).
Solution
..0.
x
Dividing the interval [0, x] into n pieces
gives ∆t =xnand ti = 0+ i∆t =
ixn. So
Rn
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 10 / 32
. . . . . .
Area as a Function
Example
Let f(t) = t3 and define g(x) =∫ x
0f(t)dt. Find g(x) and g′(x).
Solution
..0.
x
Dividing the interval [0, x] into n pieces
gives ∆t =xnand ti = 0+ i∆t =
ixn. So
Rn =xn· x
3
n3+
xn· (2x)
3
n3+ · · ·+ x
n· (nx)
3
n3
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 10 / 32
. . . . . .
Area as a Function
Example
Let f(t) = t3 and define g(x) =∫ x
0f(t)dt. Find g(x) and g′(x).
Solution
..0.
x
Dividing the interval [0, x] into n pieces
gives ∆t =xnand ti = 0+ i∆t =
ixn. So
Rn =xn· x
3
n3+
xn· (2x)
3
n3+ · · ·+ x
n· (nx)
3
n3
=x4
n4(13 + 23 + 33 + · · ·+ n3
)
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 10 / 32
. . . . . .
Area as a Function
Example
Let f(t) = t3 and define g(x) =∫ x
0f(t)dt. Find g(x) and g′(x).
Solution
..0.
x
Dividing the interval [0, x] into n pieces
gives ∆t =xnand ti = 0+ i∆t =
ixn. So
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)
]2V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 10 / 32
. . . . . .
Area as a Function
Example
Let f(t) = t3 and define g(x) =∫ x
0f(t)dt. Find g(x) and g′(x).
Solution
..0.
x
Dividing the interval [0, x] into n pieces
gives ∆t =xnand ti = 0+ i∆t =
ixn. So
Rn =x4n2(n+ 1)2
4n4
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 10 / 32
. . . . . .
Area as a Function
Example
Let f(t) = t3 and define g(x) =∫ x
0f(t)dt. Find g(x) and g′(x).
Solution
..0.
x
Dividing the interval [0, x] into n pieces
gives ∆t =xnand ti = 0+ i∆t =
ixn. So
Rn =x4n2(n+ 1)2
4n4
So g(x) = limx→∞
Rn =x4
4
and g′(x) = x3.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 10 / 32
. . . . . .
Area as a Function
Example
Let f(t) = t3 and define g(x) =∫ x
0f(t)dt. Find g(x) and g′(x).
Solution
..0.
x
Dividing the interval [0, x] into n pieces
gives ∆t =xnand ti = 0+ i∆t =
ixn. So
Rn =x4n2(n+ 1)2
4n4
So g(x) = limx→∞
Rn =x4
4and g′(x) = x3.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 10 / 32
. . . . . .
The area function in general
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 the
curve.I Question: What does f tell you about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 11 / 32
. . . . . .
Envisioning the area function
Example
Suppose f(t) is the function graphed below:
...x
..
y
....f
..2
..4
..6
..8
..10
Let g(x) =∫ x
0f(t)dt. What can you say about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32
. . . . . .
Envisioning the area function
Example
Suppose f(t) is the function graphed below:
...x
..
y
....g
....f
..2
..4
..6
..8
..10
Let g(x) =∫ x
0f(t)dt. What can you say about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32
. . . . . .
Envisioning the area function
Example
Suppose f(t) is the function graphed below:
...x
..
y
....g
....f
..2
..4
..6
..8
..10
Let g(x) =∫ x
0f(t)dt. What can you say about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32
. . . . . .
Envisioning the area function
Example
Suppose f(t) is the function graphed below:
...x
..
y
....g
....f
..2
..4
..6
..8
..10
Let g(x) =∫ x
0f(t)dt. What can you say about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32
. . . . . .
Envisioning the area function
Example
Suppose f(t) is the function graphed below:
...x
..
y
....g
....f
..2
..4
..6
..8
..10
Let g(x) =∫ x
0f(t)dt. What can you say about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32
. . . . . .
Envisioning the area function
Example
Suppose f(t) is the function graphed below:
...x
..
y
....g
....f
..2
..4
..6
..8
..10
Let g(x) =∫ x
0f(t)dt. What can you say about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32
. . . . . .
Envisioning the area function
Example
Suppose f(t) is the function graphed below:
...x
..
y
....g
....f
..2
..4
..6
..8
..10
Let g(x) =∫ x
0f(t)dt. What can you say about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32
. . . . . .
Envisioning the area function
Example
Suppose f(t) is the function graphed below:
...x
..
y
....g
....f
..2
..4
..6
..8
..10
Let g(x) =∫ x
0f(t)dt. What can you say about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32
. . . . . .
Envisioning the area function
Example
Suppose f(t) is the function graphed below:
...x
..
y
....g
....f
..2
..4
..6
..8
..10
Let g(x) =∫ x
0f(t)dt. What can you say about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32
. . . . . .
Envisioning the area function
Example
Suppose f(t) is the function graphed below:
...x
..
y
....g
....f
..2
..4
..6
..8
..10
Let g(x) =∫ x
0f(t)dt. What can you say about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32
. . . . . .
Envisioning the area function
Example
Suppose f(t) is the function graphed below:
...x
..
y
....g
....f
..2
..4
..6
..8
..10
Let g(x) =∫ x
0f(t)dt. What can you say about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32
. . . . . .
Envisioning the area function
Example
Suppose f(t) is the function graphed below:
...x
..
y
....g
....f
..2
..4
..6
..8
..10
Let g(x) =∫ x
0f(t)dt. What can you say about g?
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32
. . . . . .
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 antiderivative of f.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 13 / 32
. . . . . .
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 antiderivative of f.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 13 / 32
. . . . . .
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).
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 14 / 32
. . . . . .
Proving the Fundamental Theorem
Proof.Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=
1h
∫ 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
Somh ≤ g(x+ h)− g(x)
h≤ Mh.
As h → 0, both mh and Mh tend to f(x).
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 15 / 32
. . . . . .
Proving the Fundamental Theorem
Proof.Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=1h
∫ 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
Somh ≤ g(x+ h)− g(x)
h≤ Mh.
As h → 0, both mh and Mh tend to f(x).
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 15 / 32
. . . . . .
Proving the Fundamental Theorem
Proof.Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=1h
∫ 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
Somh ≤ g(x+ h)− g(x)
h≤ Mh.
As h → 0, both mh and Mh tend to f(x).
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 15 / 32
. . . . . .
Proving the Fundamental Theorem
Proof.Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=1h
∫ 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
Somh ≤ g(x+ h)− g(x)
h≤ Mh.
As h → 0, both mh and Mh tend to f(x).
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 15 / 32
. . . . . .
Proving the Fundamental Theorem
Proof.Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=1h
∫ 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
Somh ≤ g(x+ h)− g(x)
h≤ Mh.
As h → 0, both mh and Mh tend to f(x).
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 15 / 32
. . . . . .
Proving the Fundamental Theorem
Proof.Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=1h
∫ 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
Somh ≤ g(x+ h)− g(x)
h≤ Mh.
As h → 0, both mh and Mh tend to f(x).
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 15 / 32
. . . . . .
Proving the Fundamental Theorem
Proof.Let h > 0 be given so that x+ h < b. We have
g(x+ h)− g(x)h
=1h
∫ 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
Somh ≤ g(x+ h)− g(x)
h≤ Mh.
As h → 0, both mh and Mh tend to f(x).
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 15 / 32
. . . . . .
Meet the Mathematician: James Gregory
I Scottish, 1638-1675I Astronomer and GeometerI Conceived transcendental
numbers and foundevidence that π wastranscendental
I Proved a geometricversion of 1FTC as alemma but didn’t take itfurther
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 16 / 32
. . . . . .
Meet the Mathematician: Isaac Barrow
I English, 1630-1677I Professor of Greek,
theology, and mathematicsat Cambridge
I Had a famous student
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 17 / 32
. . . . . .
Meet the Mathematician: Isaac Newton
I English, 1643–1727I Professor at Cambridge
(England)I Philosophiae Naturalis
Principia Mathematicapublished 1687
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 18 / 32
. . . . . .
Meet the Mathematician: Gottfried Leibniz
I German, 1646–1716I Eminent philosopher as
well as mathematicianI Contemporarily disgraced
by the calculus prioritydispute
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 19 / 32
. . . . . .
Differentiation and Integration as reverse processes
Putting together 1FTC and 2FTC, we get a beautiful relationshipbetween the two fundamental concepts in calculus.
Theorem (The Fundamental Theorem(s) of Calculus)
I. If f is a continuous function, then
ddx
∫ 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.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 20 / 32
. . . . . .
Outline
Recall: The Evaluation Theorem a/k/a 2nd FTC
The First Fundamental Theorem of CalculusArea as a FunctionStatement and proof of 1FTCBiographies
Differentiation of functions defined by integrals“Contrived” examplesErfOther applications
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 21 / 32
. . . . . .
Differentiation of area functions
Example
Let h(x) =∫ 3x
0t3 dt. What is h′(x)?
Solution (Using 2FTC)
h(x) =t4
4
∣∣∣∣∣3x
0
=14(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.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 22 / 32
. . . . . .
Differentiation of area functions
Example
Let h(x) =∫ 3x
0t3 dt. What is h′(x)?
Solution (Using 2FTC)
h(x) =t4
4
∣∣∣∣∣3x
0
=14(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.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 22 / 32
. . . . . .
Differentiation of area functions
Example
Let h(x) =∫ 3x
0t3 dt. What is h′(x)?
Solution (Using 2FTC)
h(x) =t4
4
∣∣∣∣∣3x
0
=14(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.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 22 / 32
. . . . . .
Differentiation of area functions
Example
Let h(x) =∫ 3x
0t3 dt. What is h′(x)?
Solution (Using 2FTC)
h(x) =t4
4
∣∣∣∣∣3x
0
=14(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.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 22 / 32
. . . . . .
Differentiation of area functions, in general
I by 1FTCddx
∫ k(x)
af(t)dt = f(k(x))k′(x)
I by reversing the order of integration:
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)
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 23 / 32
. . . . . .
Another Example
Example
Let h(x) =∫ sin2 x
0(17t2 + 4t− 4)dt. What is h′(x)?
SolutionWe have
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
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 24 / 32
. . . . . .
Another Example
Example
Let h(x) =∫ sin2 x
0(17t2 + 4t− 4)dt. What is h′(x)?
SolutionWe have
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
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 24 / 32
. . . . . .
A Similar Example
Example
Let h(x) =∫ sin2 x
3(17t2 + 4t− 4)dt. What is h′(x)?
SolutionWe have
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
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 25 / 32
. . . . . .
A Similar Example
Example
Let h(x) =∫ sin2 x
3(17t2 + 4t− 4)dt. What is h′(x)?
SolutionWe have
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
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 25 / 32
. . . . . .
Compare
QuestionWhy is
ddx
∫ 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 derivative?
AnswerBecause∫ 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.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 26 / 32
. . . . . .
Compare
QuestionWhy is
ddx
∫ 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 derivative?
AnswerBecause∫ 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.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 26 / 32
. . . . . .
The Full Nasty
Example
Find the derivative of F(x) =∫ ex
x3sin4 t dt.
Solution
ddx
∫ ex
x3sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2
Notice here it’s much easier than finding an antiderivative for sin4.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 27 / 32
. . . . . .
The Full Nasty
Example
Find the derivative of F(x) =∫ ex
x3sin4 t dt.
Solution
ddx
∫ ex
x3sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2
Notice here it’s much easier than finding an antiderivative for sin4.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 27 / 32
. . . . . .
The Full Nasty
Example
Find the derivative of F(x) =∫ ex
x3sin4 t dt.
Solution
ddx
∫ ex
x3sin4 t dt = sin4(ex) · ex − sin4(x3) · 3x2
Notice here it’s much easier than finding an antiderivative for sin4.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 27 / 32
. . . . . .
Why use 1FTC?
QuestionWhy 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 inelementary terms.
I Some functions are naturally defined in terms of other integrals.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 28 / 32
. . . . . .
Why use 1FTC?
QuestionWhy 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 inelementary terms.
I Some functions are naturally defined in terms of other integrals.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 28 / 32
. . . . . .
Why use 1FTC?
QuestionWhy 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 inelementary terms.
I Some functions are naturally defined in terms of other integrals.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 28 / 32
. . . . . .
Erf
Here’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t2 dt.
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√πe−x2 .
Example
Findddx
erf(x2).
SolutionBy the chain rule we have
ddx
erf(x2) = erf′(x2)ddx
x2 =2√πe−(x2)22x =
4√πxe−x4 .
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 29 / 32
. . . . . .
Erf
Here’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t2 dt.
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√πe−x2 .
Example
Findddx
erf(x2).
SolutionBy the chain rule we have
ddx
erf(x2) = erf′(x2)ddx
x2 =2√πe−(x2)22x =
4√πxe−x4 .
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 29 / 32
. . . . . .
Erf
Here’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t2 dt.
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√πe−x2 .
Example
Findddx
erf(x2).
SolutionBy the chain rule we have
ddx
erf(x2) = erf′(x2)ddx
x2 =2√πe−(x2)22x =
4√πxe−x4 .
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 29 / 32
. . . . . .
Erf
Here’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t2 dt.
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√πe−x2 .
Example
Findddx
erf(x2).
SolutionBy the chain rule we have
ddx
erf(x2) = erf′(x2)ddx
x2 =2√πe−(x2)22x =
4√πxe−x4 .
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 29 / 32
. . . . . .
Erf
Here’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t2 dt.
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√πe−x2 .
Example
Findddx
erf(x2).
SolutionBy the chain rule we have
ddx
erf(x2) = erf′(x2)ddx
x2 =2√πe−(x2)22x =
4√πxe−x4 .
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 29 / 32
. . . . . .
Erf
Here’s a function with a funny name but an important role:
erf(x) =2√π
∫ x
0e−t2 dt.
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√πe−x2 .
Example
Findddx
erf(x2).
SolutionBy the chain rule we have
ddx
erf(x2) = erf′(x2)ddx
x2 =2√πe−(x2)22x =
4√πxe−x4 .
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 29 / 32
. . . . . .
Other functions defined by integrals
I The future value of an asset:
FV(t) =∫ ∞
tπ(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.
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 30 / 32
. . . . . .
Surplus by picture
..quantity (q)
.
price (p)
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 31 / 32
. . . . . .
Surplus by picture
..quantity (q)
.
price (p)
..
demand f(q)
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 31 / 32
. . . . . .
Surplus by picture
..quantity (q)
.
price (p)
..
demand f(q)
.
supply
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 31 / 32
. . . . . .
Surplus by picture
..quantity (q)
.
price (p)
..
demand f(q)
.
supply
.
equilibrium
..q∗
..
p∗
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 31 / 32
. . . . . .
Surplus by picture
..quantity (q)
.
price (p)
..
demand f(q)
.
market revenue
.
supply
.
equilibrium
..q∗
..
p∗
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 31 / 32
. . . . . .
Surplus by picture
..quantity (q)
.
price (p)
..
demand f(q)
.
market revenue
.
supply
.
equilibrium
..q∗
..
p∗
.
consumer surplus
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 31 / 32
. . . . . .
Surplus by picture
..quantity (q)
.
price (p)
..
demand f(q)
.
supply
.
equilibrium
..q∗
..
p∗
.
consumer surplus
.
producer surplus
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 31 / 32
. . . . . .
Summary
I Functions defined as integrals can be differentiated using the firstFTC:
ddx
∫ x
af(t)dt = f(x)
I The two FTCs link the two major processes in calculus:differentiation and integration∫
F′(x)dx = F(x) + C
I Follow the calculus wars on twitter: #calcwars
V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 32 / 32