lesson 24: the definite integral (section 10 version)

. . . . . .

Section5.2TheDefiniteIntegral

V63.0121, CalculusI

April15, 2009

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. . . . . .

Outline

Recall

Thedefiniteintegralasalimit

EstimatingtheDefiniteIntegral

Propertiesoftheintegral

ComparisonPropertiesoftheIntegral

. . . . . .

Cavalieri’smethodingeneralLet f beapositivefunctiondefinedontheinterval [a,b]. Wewanttofindtheareabetween x = a, x = b, y = 0, and y = f(x).Foreachpositiveinteger n, divideuptheintervalinto n pieces.

Then ∆x =b− an

. Foreach i between 1 and n, let xi bethe ith

stepbetween a and b. So

. .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

. . . . . .

FormingRiemannsumsWehavemanychoicesofrepresentativepointstoapproximatetheareaineachsubinterval.

leftendpoints…

Ln =n∑

i=1

f(xi−1)∆x

. .x. . . . . . .Ingeneral, choose ci tobeapointinthe ithinterval [xi−1, xi].Formthe Riemannsum

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

i=1

f(ci)∆x

. . . . . .


rightendpoints…

Rn =n∑

i=1

f(xi)∆x


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

i=1

f(ci)∆x

. . . . . .


midpoints…

Mn =n∑

i=1

f(xi−1 + xi

2

)∆x


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

i=1

f(ci)∆x

. . . . . .


randompoints…


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

i=1

f(ci)∆x

. . . . . .

Theoremofthe(previous)Day

TheoremIf f isacontinuousfunctionon[a,b] orhasfinitelymanyjumpdiscontinuities, then

limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}

existsandisthesamevaluenomatterwhatchoiceof ci wemade.

. .x

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . . . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . . . . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . . . . . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . . . . . . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x. . . . . . . . . . . . . . . . . . . . .

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x......................

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x.......................

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x........................

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x.........................

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x..........................

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x...........................

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x............................

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x.............................

. . . . . .



limn→∞

Sn = limn→∞

{n∑

i=1

f(ci)∆x

}


. .x..............................

. . . . . .

Outline

Recall





. . . . . .


DefinitionIf f isafunctiondefinedon [a,b], the definiteintegralof f from ato b isthenumber∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci) ∆x

. . . . . .

Notation/Terminology

∫ b

af(x)dx

I∫

— integralsign (swoopy S)

I f(x) — integrandI a and b — limitsofintegration (a isthe lowerlimit and bthe upperlimit)

I dx —??? (aparenthesis? aninfinitesimal? avariable?)I Theprocessofcomputinganintegraliscalled integration or

quadrature

. . . . . .


∫ b

af(x)dx

I∫


I f(x) — integrand

I a and b — limitsofintegration (a isthe lowerlimit and bthe upperlimit)


quadrature

. . . . . .


∫ b

af(x)dx

I∫




quadrature

. . . . . .


∫ b

af(x)dx

I∫



I dx —??? (aparenthesis? aninfinitesimal? avariable?)

I Theprocessofcomputinganintegraliscalled integration orquadrature

. . . . . .


∫ b

af(x)dx

I∫




quadrature

. . . . . .

Thelimitcanbesimplified

TheoremIf f iscontinuouson [a,b] orif f hasonlyfinitelymanyjumpdiscontinuities, then f isintegrableon [a,b]; thatis, thedefinite

integral∫ b

af(x)dx exists.

TheoremIf f isintegrableon [a,b] then∫ b

af(x)dx = lim

n→∞

n∑i=1

f(xi)∆x,

where

∆x =b− an

and xi = a + i∆x

. . . . . .

Outline

Recall





. . . . . .


Givenapartitionof [a,b] into n pieces, let x̄i bethemidpointof[xi−1, xi]. Define

Mn =n∑

i=1

f(x̄i)∆x.

. . . . . .

Example

Estimate∫ 1

0

41 + x2

dx usingthemidpointruleandfourdivisions.

SolutionThepartitionis 0 <

14

<12

<34

< 1, sotheestimateis

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

)=

150, 166,78447, 720, 465

≈ 3.1468

. . . . . .

Example

Estimate∫ 1

0

41 + x2



14

<12

<34


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

)

=150, 166,78447, 720, 465

≈ 3.1468

. . . . . .

Example

Estimate∫ 1

0

41 + x2



14

<12

<34


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

)=

150, 166,78447, 720, 465

≈ 3.1468

. . . . . .

Outline

Recall





. . . . . .


Theorem(AdditivePropertiesoftheIntegral)Let f and g beintegrablefunctionson [a,b] and c aconstant.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.

. . . . . .

MorePropertiesoftheIntegral

Conventions: ∫ a

bf(x)dx = −

∫ b

af(x)dx

∫ a

af(x)dx = 0

Thisallowsustohave

5.∫ c

af(x)dx =

∫ b

af(x)dx +

∫ c

bf(x)dx forall a, b, and c.

. . . . . .

ExampleSuppose f and g arefunctionswith

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.

. . . . . .

SolutionWehave

(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

. . . . . .

Outline

Recall





. . . . . .


TheoremLet f and g beintegrablefunctionson [a,b].

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

af(x)dx ≥ 0

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

af(x)dx ≥

∫ b

ag(x)dx

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

m(b− a) ≤∫ b

af(x)dx ≤ M(b− a)

. . . . . .

Example

Estimate∫ 2

1

1xdx usingthecomparisonproperties.

SolutionSince

12≤ x ≤ 1

1forall x in [1,2], wehave

12· 1 ≤

∫ 2

1

1xdx ≤ 1 · 1

lesson 24: the definite integral (section 10 version)

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