Lesson 25: The Fundamental Theorem of Calculus

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Section 5.4The Fundamental Theorem of Calculus

V63.0121.006/016, Calculus I

New York University

April 22, 2010

Announcements

I April 29: Movie DayI April 30: Quiz 5 on §§5.1–5.4I Monday, May 10, 12:00noon Final Exam

. . . . . .

. . . . . .

Announcements

I April 29: Movie DayI April 30: Quiz 5 on

§§5.1–5.4I 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

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Resurrection policies

I Current distribution of grade: 40% final, 25% midterm, 15%quizzes, 10% written HW, 10% WebAssign

I Remember we drop the lowest quiz, lowest written HW, and5 lowest WebAssign-ments

I If your final exam score beats your midterm score, we willre-weight it by 50% and make the midterm 15%


. . . . . .

Resurrection policies

I Current distribution of grade: 40% final, 25% midterm, 15%quizzes, 10% written HW, 10% WebAssign

I Remember we drop the lowest quiz, lowest written HW, and5 lowest WebAssign-ments

I If your final exam score beats your midterm score, we willre-weight it by 50% and make the midterm 15%


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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.


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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


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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


. . . . . .

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).


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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:


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TheoremIf v(t) represents the velocity of a particle moving rectilinearly, then∫ t1

t0v(t)dt = s(t1)− s(t0).


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TheoremIf MC(x) represents the marginal cost of making x units of a product,then

C(x) = C(0) +∫ x

0MC(q)dq.


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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.


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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


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Outline





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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 =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)

]2=

x4n2(n+ 1)2

4n4→ x4

4as n → ∞.


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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 =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)

]2=

x4n2(n+ 1)2

4n4→ x4

4as n → ∞.


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An area function, continued

So

g(x) =x4

4.

This means thatg′(x) = x3.


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An area function, continued

So

g(x) =x4

4.

This means thatg′(x) = x3.


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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 the

curve.I Question: What does f tell you about g?


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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?


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Example


. ..x

..y

.

..

.g

.

.. .f

..2

..4

..6

..8

..10

Let g(x) =∫ x



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Example


. ..x

..y

.

..

.g

.

.. .f

..2

..4

..6

..8

..10

Let g(x) =∫ x



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Example


. ..x

..y

.

..

.g

.

.. .f

..2

..4

..6

..8

..10

Let g(x) =∫ x



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Example


. ..x

..y

.

..

.g

.

.. .f

..2

..4

..6

..8

..10

Let g(x) =∫ x



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Example


. ..x

..y

.

..

.g

.

.. .f

..2

..4

..6

..8

..10

Let g(x) =∫ x



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Example


. ..x

..y

.

..

.g

.

.. .f

..2

..4

..6

..8

..10

Let g(x) =∫ x



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Example


. ..x

..y

.

..

.g

.

.. .f

..2

..4

..6

..8

..10

Let g(x) =∫ x



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Example


. ..x

..y

.

..

.g

.

.. .f

..2

..4

..6

..8

..10

Let g(x) =∫ x



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Example


. ..x

..y

.

..

.g

.

.. .f

..2

..4

..6

..8

..10

Let g(x) =∫ x



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Example


. ..x

..y

.

..

.g

.

.. .f

..2

..4

..6

..8

..10

Let g(x) =∫ x



. . . . . .


Example


. ..x

..y

.

..

.g

.

.. .f

..2

..4

..6

..8

..10

Let g(x) =∫ x



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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.


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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.


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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).


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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).


. . . . . .



g(x+ h)− g(x)h

=1h

∫ x+h

xf(t)dt.


mh · h ≤

∫ x+h

xf(t)dt

≤ Mh · h


h≤ Mh.



. . . . . .



g(x+ h)− g(x)h

=1h

∫ x+h

xf(t)dt.


mh · h ≤

∫ x+h

xf(t)dt

≤ Mh · h


h≤ Mh.



. . . . . .



g(x+ h)− g(x)h

=1h

∫ x+h

xf(t)dt.


mh · h ≤

∫ x+h

xf(t)dt ≤ Mh · h


h≤ Mh.



. . . . . .



g(x+ h)− g(x)h

=1h

∫ x+h

xf(t)dt.


mh · h ≤∫ x+h

xf(t)dt ≤ Mh · h


h≤ Mh.



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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


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Meet the Mathematician: Isaac Barrow

I English, 1630-1677I Professor of Greek,

theology, and mathematicsat Cambridge

I Had a famous student


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Meet the Mathematician: Isaac Newton

I English, 1643–1727I Professor at Cambridge

(England)I Philosophiae Naturalis

Principia Mathematicapublished 1687


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Meet the Mathematician: Gottfried Leibniz

I German, 1646–1716I Eminent philosopher as

well as mathematicianI Contemporarily disgraced

by the calculus prioritydispute


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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.


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Outline





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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.


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.


. . . . . .


Example

Let h(x) =∫ 3x



h(x) =t4

4

∣∣∣∣∣3x

0

=14(3x)4 = 1

4 · 81x4, so h′(x) = 81x3.



0t3 dt and


h′(x) = g′(k(x)) · k′(x) = (k(x))3 · 3 = (3x)3 · 3 = 81x3.


. . . . . .


Example

Let h(x) =∫ 3x



h(x) =t4

4

∣∣∣∣∣3x

0

=14(3x)4 = 1

4 · 81x4, so h′(x) = 81x3.



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.


. . . . . .


Example

Let h(x) =∫ 3x



h(x) =t4

4

∣∣∣∣∣3x

0

=14(3x)4 = 1

4 · 81x4, so h′(x) = 81x3.



0t3 dt and


h′(x) = g′(k(x)) · k′(x) = (k(x))3 · 3 = (3x)3 · 3 = 81x3.


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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)


. . . . . .

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


. . . . . .

Example


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


. . . . . .

Example


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


. . . . . .

Example


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


. . . . . .

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.


. . . . . .

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.


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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.


. . . . . .

Example


x3sin4 t dt.

Solution

ddx

∫ ex




. . . . . .

Example


x3sin4 t dt.

Solution

ddx

∫ ex




. . . . . .

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.


. . . . . .

Why use 1FTC?


Answer




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Why use 1FTC?


Answer




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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 .


. . . . . .

Erf


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),


Example

Findddx

erf(x2).


ddx


x2 =2√πe−(x2)22x =

4√πxe−x4 .


. . . . . .

Erf


erf(x) =2√π

∫ x

0e−t2 dt.


explicitly, but we do know its derivative: erf′(x) =

2√πe−x2 .

Example

Findddx

erf(x2).


ddx


x2 =2√πe−(x2)22x =

4√πxe−x4 .


. . . . . .

Erf


erf(x) =2√π

∫ x

0e−t2 dt.



Example

Findddx

erf(x2).


ddx


x2 =2√πe−(x2)22x =

4√πxe−x4 .


. . . . . .

Erf


erf(x) =2√π

∫ x

0e−t2 dt.



Example

Findddx

erf(x2).


ddx


x2 =2√πe−(x2)22x =

4√πxe−x4 .


. . . . . .

Erf


erf(x) =2√π

∫ x

0e−t2 dt.



Example

Findddx

erf(x2).


ddx


x2 =2√πe−(x2)22x =

4√πxe−x4 .


. . . . . .

Other functions defined by integrals

I The future value of an asset:

FV(t) =∫ ∞

tπ(τ)e−rτ dτ

where π(τ) is the profitability at time τ 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.


. . . . . .

Surplus by picture

..quantity (q)

.price (p)

.

.demand f(q)

.supply

.equilibrium

..q∗

..p∗

.consumer surplus


. . . . . .

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


http://twitter.com/#search?q=%23calcwars

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Lesson 25: The Fundamental Theorem of Calculus