Lesson 24: Areas and Distances, The Definite Integral (handout)

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Sec on 5.1–5.2Areas and Distances, The Definite

Integral

V63.0121.001: Calculus IProfessor Ma hew Leingang

New York University

April 25, 2011

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Announcements

I Quiz 5 on Sec ons4.1–4.4 April 28/29

I Final Exam Thursday May12, 2:00–3:50pm

I cumula veI loca on TBDI old exams on common

website

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Objectives from Section 5.1I Compute the area of a region byapproxima ng it with rectanglesand le ng the size of therectangles tend to zero.

I Compute the total distancetraveled by a par cle byapproxima ng it as distance =(rate)( me) and le ng the meintervals over which oneapproximates tend to zero.

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Notes

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Notes

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Notes

. 1.

. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011

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Objectives from Section 5.2

I Compute the definite integralusing a limit of Riemann sums

I Es mate the definite integralusing a Riemann sum (e.g.,Midpoint Rule)

I Reason with the definite integralusing its elementary proper es.

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OutlineArea through the Centuries

EuclidArchimedesCavalieri

Generalizing Cavalieri’s methodAnalogies

DistancesOther applica ons

The definite integral as a limitEs ma ng the Definite IntegralProper es of the integralComparison Proper es of the Integral

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Easy Areas: RectangleDefini onThe area of a rectangle with dimensions ℓ and w is the productA = ℓw.

..ℓ

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w

It may seem strange that this is a defini on and not a theorem butwe have to start somewhere.

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Easy Areas: ParallelogramBy cu ng and pas ng, a parallelogram can be made into a rectangle.

..b

.b

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h

SoFactThe area of a parallelogram of base width b and height h is

A = bh

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Easy Areas: TriangleBy copying and pas ng, a triangle can be made into a parallelogram.

..b

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h

SoFactThe area of a triangle of base width b and height h is

A =12bh

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Easy Areas: Other PolygonsAny polygon can be triangulated, so its area can be found bysumming the areas of the triangles:

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Hard Areas: Curved Regions

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???

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Meet the mathematician: Archimedes

I Greek (Syracuse), 287 BC– 212 BC (a er Euclid)

I GeometerI Weapons engineer

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Archimedes and the Parabola

..

1

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18

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18

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164

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164

.164

.164

Archimedes found areas of a sequence of triangles inscribed in aparabola.

A = 1+ 2 · 18+ 4 · 1

64+ · · · = 1+

14+

116

+ · · ·+ 14n

+ · · ·

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Summing a geometric seriesFactFor any number r and any posi ve integer n,

(1− r)(1+ r+ r2 + · · ·+ rn) = 1− rn+1.

Proof.

(1− r)(1+ r+ r2 + · · ·+ rn)= (1+ r+ r2 + · · ·+ rn)− r(1+ r+ r2 + · · ·+ rn)= (1+ r+ r2 + · · ·+ rn)− (r+ r2 + r3 · · ·+ rn + rn+1)

= 1− rn+1

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Summing a geometric seriesFactFor any number r and any posi ve integer n,

(1− r)(1+ r+ r2 + · · ·+ rn) = 1− rn+1.

Corollary

1+ r+ · · ·+ rn =1− rn+1

1− r

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Summing the series

We need to know the value of the series

1+14+

116

+ · · ·+ 14n

+ · · ·

Using the corollary,

1+14+

116

+ · · ·+ 14n

=1− (1/4)n+1

1− 1/4→ 1

3/4=

43as n → ∞.

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Notes

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Cavalieri

I Italian,1598–1647

I Revisitedthe areaproblemwith adifferentperspec ve

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Cavalieri’s method

..

y = x2

..0..

1

..12

..

Divide up the interval into pieces andmeasure the area of the inscribedrectangles:

L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =1

125+

4125

+9

125+

16125

=30125

Ln =?

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The Square Pyramidial Numbers

FactLet n be a posi ve integer. Then

1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)

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This formula was known to the Arabs and discussed by Fibonacci inhis book Liber Abaci.

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What is Ln?Divide the interval [0, 1] into n pieces. Then each has width

1n. The

rectangle over the ith interval and under the parabola has area

1n·(i− 1n

)2

=(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=

1+ 22 + 32 + · · ·+ (n− 1)2

n3

SoLn =

n(n− 1)(2n− 1)6n3

→ 13

as n → ∞.

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Cavalieri’s method for different functionsTry the same trick with f(x) = x3. We have

Ln =1n· f(1n

)+

1n· f(2n

)+ · · ·+ 1

n· f(n− 1n

)=

1n· 1n3

+1n· 2

3

n3+ · · ·+ 1

n· (n− 1)3

n3

=1+ 23 + 33 + · · ·+ (n− 1)3

n4

=n2(n− 1)2

4n4→ 1

4as n → ∞.

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Nicomachus’s Theorem

Fact (Nicomachus 1st c. CE, Aryabhata 5th c., Al-Karaji 11th c.)

1+ 23 + 33 + · · ·+ (n− 1)3 = [1+ 2+ · · ·+ (n− 1)]2

=[12n(n− 1)

]2

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Cavalieri’s method with different heights

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Rn =1n· 1

3

n3+

1n· 2

3

n3+ · · ·+ 1

n· n

3

n3

=13 + 23 + 33 + · · ·+ n3

n4

=1n4

[12n(n+ 1)

]2=

n2(n+ 1)2

4n4→ 1

4as n → ∞.

So even though the rectangles overlap, we s ll get the same answer.

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Cavalieri’s method in generalProblem

.. x..x0..x1

..xi

..xn−1

..xn

.. . .

.. . .

Let f be a posi ve func on definedon the interval [a, b]. Find thearea between x = a, x = b, y = 0,and y = f(x).

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Cavalieri’s method in generalFor each posi ve integer n, divide up the interval into n pieces. Then∆x =

b− an

. For each i between 1 and n, let xi be the ith stepbetween a and b.

.. x..x0..x1

..xi

..xn−1

..xn

.. . .

.. . .

x0 = a

x1 = x0 +∆x = a+b− an

x2 = x1 +∆x = a+ 2 · b− an

. . .

xi = a+ i · b− an

. . .

xn = a+ n · b− an

= b

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Forming Riemann Sums

Choose ci to be a point in the ith interval [xi−1, xi]. Form theRiemann sum

Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑

i=1

f(ci)∆x

Thus we approximate area under a curve by a sum of areas ofrectangles.

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Forming Riemann sumsWe have many choices of representa ve points to approximate thearea in each subinterval.

…even random points!

.. x.......

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Theorem of the DayTheoremIf f is a con nuous func on on[a, b] or has finitely many jumpdiscon nui es, then

limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}

exists and is the same value noma er what choice of ci we make.

.... x.

M15 = 7.49968

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AnalogiesThe Tangent Problem(Ch. 2–4)

I Want the slope of a curveI Only know the slope oflines

I Approximate curve with aline

I Take limit over be er andbe er approxima ons

The Area Problem (Ch. 5)I Want the area of a curvedregion

I Only know the area ofpolygons

I Approximate region withpolygons

I Take limit over be er andbe er approxima ons

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Distances

Just like area = length× width, we have

distance = rate× me.

So here is another use for Riemann sums.

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Application: Dead Reckoning

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Computing position by Dead ReckoningExampleA sailing ship is cruising back and forth along a channel (in a straightline). At noon the ship’s posi on and velocity are recorded, butshortly therea er a storm blows in and posi on is impossible tomeasure. The velocity con nues to be recorded at thirty-minuteintervals.

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Computing position by Dead ReckoningExample

Time 12:00 12:30 1:00 1:30 2:00Speed (knots) 4 8 12 6 4Direc on E E E E WTime 2:30 3:00 3:30 4:00Speed 3 3 5 9Direc on W E E E

Es mate the ship’s posi on at 4:00pm.

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SolutionSolu onWe es mate that the speed of 4 knots (nau cal miles per hour) ismaintained from 12:00 un l 12:30. So over this me interval theship travels (

4 nmihr

)(12hr)

= 2 nmi

We can con nue for each addi onal half hour and get

distance = 4× 1/2 + 8× 1/2 + 12× 1/2

+ 6× 1/2 − 4× 1/2 − 3× 1/2 + 3× 1/2 + 5× 1/2 = 15.5

So the ship is 15.5 nmi east of its original posi on.

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Analysis

I This method of measuring posi on by recording velocity wasnecessary un l global-posi oning satellite technology becamewidespread

I If we had velocity es mates at finer intervals, we’d get be eres mates.

I If we had velocity at every instant, a limit would tell us ourexact posi on rela ve to the last me we measured it.

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Other uses of Riemann sums

Anything with a product!I Area, volumeI Anything with a density: Popula on, massI Anything with a “speed:” distance, throughput, powerI Consumer surplusI Expected value of a random variable

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The definite integral as a limit

Defini 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

∆x→0

n∑i=1

f(ci)∆x

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Notation/Terminology∫ b

af(x) dx = lim

∆x→0

n∑i=1

f(ci)∆x

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∫— integral sign (swoopy S)

I f(x)— integrandI a and b— limits of integra on (a is the lower limit and b theupper limit)

I dx— ??? (a parenthesis? an infinitesimal? a variable?)I The process of compu ng an integral is called integra on orquadrature

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The limit can be simplified

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.

So we can find the integral by compu ng the limit of any sequenceof Riemann sums that we like,

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Example

Find∫ 3

0x dx

Solu on

For any n we have∆x =3nand for each i between 0 and n, xi =

3in.

For each i, take xi to represent the func on on the ith interval. So∫ 3

0x dx = lim

n→∞Rn = lim

n→∞

n∑i=1

f(xi)∆x = limn→∞

n∑i=1

(3in

) (3n

)= lim

n→∞

9n2

n∑i=1

i = limn→∞

9n2

· n(n+ 1)2

=92· 1

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. 14.


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Example

Find∫ 3

0x2 dx

Solu on

For any n and i we have∆x =3nand xi =

3in. So∫ 3

0x2 dx = lim

x→∞Rn = lim

x→∞

n∑i=1

f(xi)∆x = limx→∞

n∑i=1

(3in

)2(3n

)= lim

x→∞

27n3

n∑i=1

i2 = limx→∞

27n3

· n(n+ 1)(2n+ 1)6

=273

= 9

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Example

Find∫ 3

0x3 dx

Solu on

For any n we have∆x =3nand xi =

3in. So

Rn =n∑

i=1

f(xi)∆x =n∑

i=1

(3in

)3(3n

)=

81n4

n∑i=1

i3

=81n4

· n2(n+ 1)2

4−→ 81

4

So∫ 3

0x3 dx =

814

= 20.25

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Estimating the Definite IntegralExample

Es mate∫ 1

0

41+ x2

dx usingM4.

Solu on

We have x0 = 0, x1 =14, x2 =

12, x3 =

34, x4 = 1.

So c1 =18, c2 =

38, c3 =

58, c4 =

78.

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Es mate∫ 1

0

41+ x2

dx usingM4.

Solu on

M4 =14

(4

1+ (1/8)2+

41+ (3/8)2

+4

1+ (5/8)2+

41+ (7/8)2

)=

14

(4

65/64+

473/64

+4

89/64+

4113/64

)=

6465

+6473

+6489

+64113

≈ 3.1468

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Es mate∫ 1

0

41+ x2

dx using L4 and R4

Answer

L4 =14

(4

1+ (0)2+

41+ (1/4)2

+4

1+ (1/2)2+

41+ (3/4)2

)= 1+

1617

+45+

1625

≈ 3.38118

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. 16.


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Es mate∫ 1

0

41+ x2

dx using L4 and R4

Answer

R4 =14

(4

1+ (1/4)2+

41+ (1/2)2

+4

1+ (3/4)2+

41+ (1)2

)=

1617

+45+

1625

+12≈ 2.88118

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Properties of the integralTheorem (Addi ve Proper es of the Integral)

Let f and g be integrable func ons on [a, b] and c a constant. Then

1.∫ b

ac dx = c(b− a)

2.∫ b

a[f(x) + g(x)] dx =

∫ b

af(x) dx+

∫ b

ag(x) dx.

3.∫ b

acf(x) dx = c

∫ b

af(x) dx.

4.∫ b

a[f(x)− g(x)] dx =

∫ b

af(x) dx−

∫ b

ag(x) dx.

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. 17.


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ProofsProofs.

I When integra ng a constant func on c, each Riemann sumequals c(b− a).

I A Riemann sum for f+ g equals a Riemann sum for f plus aRiemann sum for g. Using the sum rule for limits, the integralof a sum is the sum of the integrals.

I Di o for constant mul plesI Di o for differences

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Example

Find∫ 3

0

(x3 − 4.5x2 + 5.5x+ 1

)dx

Solu on

∫ 3

0(x3−4.5x2 + 5.5x+ 1) dx

=

∫ 3

0x3 dx− 4.5

∫ 3

0x2 dx+ 5.5

∫ 3

0x dx+

∫ 3

01 dx

= 20.25− 4.5 · 9+ 5.5 · 4.5+ 3 · 1 = 7.5

(This is the func on we were es ma ng the integral of before)

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More Properties of the IntegralConven ons: ∫ a

bf(x) dx = −

∫ b

af(x) dx∫ a

af(x) dx = 0

This allows us to haveTheorem

5.∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c

bf(x) dx for all a, b, and c.

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. 18.


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Illustrating Property 5Theorem

5.∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c


..x

.

y

..a

..b

..c

.

∫ b

af(x) dx

.

∫ c

bf(x) dx

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Illustrating Property 5Theorem

5.∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c


..x

.

y

..a..

b..

c.

∫ c

bf(x) dx =

−∫ b

cf(x) dx

.

∫ c

af(x) dx

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Using the PropertiesExampleSuppose f and g are func onswith

I

∫ 4

0f(x) dx = 4

I

∫ 5

0f(x) dx = 7

I

∫ 5

0g(x) dx = 3.

Find

(a)∫ 5

0[2f(x)− g(x)] dx

(b)∫ 5

4f(x) dx.

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Solu on

We have

(a)

∫ 5

0[2f(x)− g(x)] dx = 2

∫ 5

0f(x) dx−

∫ 5

0g(x) dx

= 2 · 7− 3 = 11

(b)

∫ 5

4f(x) dx =

∫ 5

0f(x) dx−

∫ 4

0f(x) dx

= 7− 4 = 3

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Comparison Properties of the IntegralTheoremLet f and g be integrable func ons on [a, b].

6. If f(x) ≥ 0 for all x in [a, b], then∫ b

af(x) dx ≥ 0

7. If f(x) ≥ g(x) for all x in [a, b], then∫ b

af(x) dx ≥

∫ b

ag(x) dx

8. If m ≤ f(x) ≤ M for all x in [a, b], then

m(b− a) ≤∫ b

af(x) dx ≤ M(b− a)

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. 20.


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Integral of a nonnegative function is nonnegativeProof.If f(x) ≥ 0 for all x in [a, b], then forany number of divisions n and choiceof sample points {ci}:

Sn =n∑

i=1

f(ci)︸︷︷︸≥0

∆x ≥n∑

i=1

0 ·∆x = 0

.. x.......Since Sn ≥ 0 for all n, the limit of {Sn} is nonnega ve, too:∫ b

af(x) dx = lim

n→∞Sn︸︷︷︸≥0

≥ 0

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The integral is “increasing”Proof.Let h(x) = f(x)− g(x). If f(x) ≥ g(x)for all x in [a, b], then h(x) ≥ 0 for allx in [a, b]. So by the previousproperty ∫ b

ah(x) dx ≥ 0 .. x.

f(x)

.

g(x)

.

h(x)

This means that∫ b

af(x) dx−

∫ b

ag(x) dx =

∫ b

a(f(x)− g(x)) dx =

∫ b

ah(x) dx ≥ 0

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Bounding the integralProof.Ifm ≤ f(x) ≤ M on for all x in [a, b], then bythe previous property∫ b

amdx ≤

∫ b

af(x) dx ≤

∫ b

aMdx

By Property 8, the integral of a constantfunc on is the product of the constant andthe width of the interval. So:

m(b− a) ≤∫ b

af(x) dx ≤ M(b− a)

.. x.

y

.

M

.

f(x)

.

m

..a

..b

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. 21.


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Example

Es mate∫ 2

1

1xdx using the comparison proper es.

Solu on

Since12≤ x ≤ 1

1for all x in [1, 2], we have

12· 1 ≤

∫ 2

1

1xdx ≤ 1 · 1

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Summary

I We can compute the area of a curved region with a limit ofRiemann sums

I We can compute the distance traveled from the velocity with alimit of Riemann sums

I Many other important uses of this process.

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Summary

I The definite integral is a limit of Riemann SumsI The definite integral can be es mated with Riemann SumsI The definite integral can be distributed across sums andconstant mul ples of func ons

I The definite integral can be bounded using bounds for thefunc on

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. 22.


Lesson 24: Areas and Distances, The Definite Integral (handout)

Technology

n n2 n

n nn nn n

n nn n nn3

n pieces

n n nn3 sonn

n4 n2 n

integral noteseasy areas

posi ve integer n