lesson 26: the fundamental theorem of calculus (handout)

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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

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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

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ObjectivesI State and explain theFundemental Theoremsof Calculus

I Use the first fundamentaltheorem of calculus tofind deriva ves offunc ons defined asintegrals.

I Compute the averagevalue of an integrablefunc on over a closedinterval.

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. Sec on 5.4: The Fundamental Theorem. V63.0121.001: Calculus I . May 2, 2011

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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

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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].

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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.

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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).

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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).

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Corollary

If 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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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.

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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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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 =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

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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 =x4n2(n+ 1)2

4n4

So g(x) = limx→∞

Rn =x4

4and g′(x) = x3.

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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?

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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

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Features of g from f

...x

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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.

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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).

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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).

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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).

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Meet the Mathematician: JamesGregory

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

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Meet the Mathematician: IsaacBarrow

I English, 1630-1677I Professor of Greek, theology,and mathema cs at Cambridge

I Had a famous student

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Meet the Mathematician: IsaacNewton

I English, 1643–1727I Professor at Cambridge(England)

I Philosophiae Naturalis PrincipiaMathema ca published 1687

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Meet the Mathematician: GottfriedLeibniz

I German, 1646–1716I Eminent philosopher as well asmathema cian

I Contemporarily disgraced by thecalculus priority dispute

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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.

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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.

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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.


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.

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Differentiation of area functionsExample

Let h(x) =∫ 3x

0t3 dt. What is h′(x)?


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.

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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)

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Another ExampleExample

Let h(x) =∫ sin2 x

0(17t2 + 4t− 4) dt. What is h′(x)?

Solu on

We 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

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A Similar ExampleExample

Let h(x) =∫ sin2 x

3(17t2 + 4t− 4) dt. What is h′(x)?

Solu on

We 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

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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.

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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.

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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.

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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

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erf(x)

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Example of erfExample

Findddx

erf(x2).

Solu on

By the chain rule we have

ddx

erf(x2) = erf′(x2)ddx

x2 =2√πe−(x2)22x =

4√πxe−x4.

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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.

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Surplus by picture

..quan ty (q)

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price (p)

..

demand f(q).

market revenue

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supply

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equilibrium

..q∗

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p∗

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consumer surplus

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producer surplus

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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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lesson 26: the fundamental theorem of calculus (handout)

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