lesson 24: areas, distances, the integral (section 021 slides

..

Sections 5.1–5.2Areas and DistancesThe Definite Integral

V63.0121.021, Calculus I

New York University

December 2, 2010

Announcements

I Final December 20, 12:00–1:50pm

. . . . . .

. . . . . .

Announcements

I Final December 20,12:00–1:50pm

I cumulativeI location TBDI old exams on commonwebsite

V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 2 / 56

. . . . . .

Objectives from Section 5.1

I Compute the area of aregion by approximating itwith rectangles and lettingthe size of the rectanglestend to zero.

I Compute the total distancetraveled by a particle byapproximating it asdistance = (rate)(time) andletting the time intervalsover which oneapproximates tend to zero.


. . . . . .

Objectives from Section 5.2

I Compute the definiteintegral using a limit ofRiemann sums

I Estimate the definiteintegral using a Riemannsum (e.g., Midpoint Rule)

I Reason with the definiteintegral using itselementary properties.


. . . . . .

Outline

Area through the CenturiesEuclidArchimedesCavalieri

Generalizing Cavalieri’s methodAnalogies

DistancesOther applications

The definite integral as a limit

Estimating the Definite Integral

Properties of the integral

Comparison Properties of the Integral


. . . . . .

Easy Areas: Rectangle

DefinitionThe area of a rectangle with dimensions ℓ and w is the product A = ℓw.

..ℓ

.

w

It may seem strange that this is a definition and not a theorem but wehave to start somewhere.


. . . . . .

Easy Areas: Parallelogram

By cutting and pasting, a parallelogram can be made into a rectangle.

..b

.b

.

h

So

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

A = bh


. . . . . .

Easy Areas: Parallelogram

By cutting and pasting, a parallelogram can be made into a rectangle.

.

.b

.b

.

h

So

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

A = bh


. . . . . .

Easy Areas: Triangle

By copying and pasting, a triangle can be made into a parallelogram.

..b

.

h

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

A =12bh


. . . . . .

Easy Areas: Other Polygons

Any polygon can be triangulated, so its area can be found by summingthe areas of the triangles:

.

.


. . . . . .

Hard Areas: Curved Regions

.

.

???


. . . . . .

Meet the mathematician: Archimedes

I Greek (Syracuse), 287 BC– 212 BC (after Euclid)

I GeometerI Weapons engineer


. . . . . .

Archimedes and the Parabola

.

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

A =

1+ 2 · 18+ 4 · 1

64+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·


. . . . . .


..

1


A = 1

+ 2 · 18+ 4 · 1

64+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·


. . . . . .


..

1

.

18

.

18


A = 1+ 2 · 18

+ 4 · 164

+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·


. . . . . .


..

1

.

18

.

18

.

164

.

164

.

164

.

164


A = 1+ 2 · 18+ 4 · 1

64+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·


. . . . . .

Summing the series

We would then need to know the value of the series

1+14+

116

+ · · ·+ 14n

+ · · ·

FactFor any number r and any positive integer n,

(1− r)(1+ r+ · · ·+ rn) = 1− rn+1

So

1+ r+ · · ·+ rn =1− rn+1

1− r

Therefore

1+14+

116

+ · · ·+ 14n

=1− (1/4)n+1

1− 1/4→ 1

3/4=43as n → ∞.


. . . . . .

Summing the series


1+14+

116

+ · · ·+ 14n

+ · · ·


(1− r)(1+ r+ · · ·+ rn) = 1− rn+1

So

1+ r+ · · ·+ rn =1− rn+1

1− r

Therefore

1+14+

116

+ · · ·+ 14n

=1− (1/4)n+1

1− 1/4

→ 13/4

=43as n → ∞.


. . . . . .

Summing the series


1+14+

116

+ · · ·+ 14n

+ · · ·


(1− r)(1+ r+ · · ·+ rn) = 1− rn+1

So

1+ r+ · · ·+ rn =1− rn+1

1− r

Therefore

1+14+

116

+ · · ·+ 14n

=1− (1/4)n+1

1− 1/4→ 1

3/4=43as n → ∞.


. . . . . .

Cavalieri

I Italian,1598–1647

I Revisited theareaproblem witha differentperspective


. . . . . .

Cavalieri's method

..

y = x2

..0..

1

..12

Divide up the interval intopieces and measure the area ofthe inscribed rectangles:

L2 =18

L3 =

127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

..

y = x2

..0..

1..

12


L2 =18

L3 =

127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

..

y = x2

..0..

1

..12

..13

..23


L2 =18

L3 =

127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

..

y = x2

..0..

1

..12

..13

..23


L2 =18

L3 =127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

..

y = x2

..0..

1

..12

..14

..24

..34


L2 =18

L3 =127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

..

y = x2

..0..

1

..12

..14

..24

..34


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

..

y = x2

..0..

1

..12

..15

..25

..35

..45


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =

1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

..

y = x2

..0..

1

..12

..15

..25

..35

..45


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

Cavalieri's method

..

y = x2

..0..

1

..12

..


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =1125

+4125

+9125

+16125

=30125

Ln =?


. . . . . .

What is Ln?

Divide the interval [0,1] into n pieces. Then each has width1n.

Therectangle 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

The Arabs knew that

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

6So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.


. . . . . .

What is Ln?

Divide the interval [0,1] into n pieces. Then each has width1n. 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

The Arabs knew that

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

6So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.


. . . . . .

What is Ln?



1n·(i− 1n

)2=

(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=1+ 22 + 32 + · · ·+ (n− 1)2

n3

The Arabs knew that

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

6So

Ln =n(n− 1)(2n− 1)

6n3

→ 13

as n → ∞.


. . . . . .

What is Ln?



1n·(i− 1n

)2=

(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=1+ 22 + 32 + · · ·+ (n− 1)2

n3

The Arabs knew that

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

6So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56

. . . . . .

Cavalieri's method for different functions

Try 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

The formula out of the hat is

1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.


. . . . . .



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


1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.


. . . . . .



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


1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2

So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.


. . . . . .



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


1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56

. . . . . .

Cavalieri's method with different heights

.

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 still get the same answer.


. . . . . .

Outline









. . . . . .

Cavalieri's method in general

Let f be a positive function defined on the interval [a,b]. We want tofind the area between x = a, x = b, y = 0, and y = f(x).For each positive 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 step betweena 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


. . . . . .

Forming Riemann sums

We have many choices of representative points to approximate thearea in each subinterval.

left endpoints…

Ln =n∑

i=1

f(xi−1)∆x

.. x.......

In general, 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


. . . . . .



right endpoints…

Rn =n∑

i=1

f(xi)∆x

.. 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

.. x.......


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

i=1

f(ci)∆x


. . . . . .



the minimum value on theinterval…

.. x.......


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

i=1

f(ci)∆x


. . . . . .



the maximum value on theinterval…

.. x.......


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

i=1

f(ci)∆x


. . . . . .



…even random points!

.. x.......


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

i=1

f(ci)∆x


. . . . . .



…even random points!

.. x.......In general, 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


. . . . . .

Theorem of the Day

Theorem

If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then

limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}

exists and is the same value nomatter what choice of ci we make. .... x.


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L1 = 3.0


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L2 = 5.25


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L3 = 6.0


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L4 = 6.375


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L5 = 6.59988


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L6 = 6.75


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L7 = 6.85692


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L8 = 6.9375


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L9 = 6.99985


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L10 = 7.04958


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L11 = 7.09064


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L12 = 7.125


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L13 = 7.15332


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L14 = 7.17819


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L15 = 7.19977


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L16 = 7.21875


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L17 = 7.23508


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L18 = 7.24927


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L19 = 7.26228


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L20 = 7.27443


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L21 = 7.28532


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L22 = 7.29448


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L23 = 7.30406


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L24 = 7.3125


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L25 = 7.31944


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L26 = 7.32559


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L27 = 7.33199


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L28 = 7.33798


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L29 = 7.34372


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


left endpoints.

L30 = 7.34882


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R1 = 12.0


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R2 = 9.75


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R3 = 9.0


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R4 = 8.625


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R5 = 8.39969


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R6 = 8.25


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R7 = 8.14236


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R8 = 8.0625


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R9 = 7.99974


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R10 = 7.94933


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R11 = 7.90868


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R12 = 7.875


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R13 = 7.84541


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R14 = 7.8209


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R15 = 7.7997


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R16 = 7.78125


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R17 = 7.76443


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R18 = 7.74907


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R19 = 7.73572


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R20 = 7.7243


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R21 = 7.7138


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R22 = 7.70335


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R23 = 7.69531


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R24 = 7.6875


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R25 = 7.67934


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R26 = 7.6715


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R27 = 7.66508


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R28 = 7.6592


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R29 = 7.65388


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


right endpoints.

R30 = 7.64864


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M1 = 7.5


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M2 = 7.5


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M3 = 7.5


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M4 = 7.5


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M5 = 7.4998


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M6 = 7.5


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M7 = 7.4996


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M8 = 7.5


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M9 = 7.49977


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M10 = 7.49947


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M11 = 7.49966


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M12 = 7.5


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M13 = 7.49937


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M14 = 7.49954


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M15 = 7.49968


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M16 = 7.49988


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M17 = 7.49974


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M18 = 7.49916


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M19 = 7.49898


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M20 = 7.4994


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M21 = 7.49951


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M22 = 7.49889


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M23 = 7.49962


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M24 = 7.5


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M25 = 7.49939


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M26 = 7.49847


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M27 = 7.4985


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M28 = 7.4986


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M29 = 7.49878


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M30 = 7.49872


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U1 = 12.0


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U2 = 10.55685


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U3 = 10.0379


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U4 = 9.41515


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U5 = 8.96004


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U6 = 8.76895


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U7 = 8.6033


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U8 = 8.45757


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U9 = 8.34564


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U10 = 8.27084


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U11 = 8.20132


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U12 = 8.13838


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U13 = 8.0916


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U14 = 8.05139


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U15 = 8.01364


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U16 = 7.98056


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U17 = 7.9539


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U18 = 7.92815


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U19 = 7.90414


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U20 = 7.88504


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U21 = 7.86737


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U22 = 7.84958


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U23 = 7.83463


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U24 = 7.82187


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U25 = 7.80824


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U26 = 7.79504


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U27 = 7.78429


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U28 = 7.77443


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U29 = 7.76495


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


maximum points.

U30 = 7.7558


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L1 = 3.0


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L2 = 4.44312


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L3 = 4.96208


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L4 = 5.58484


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L5 = 6.0395


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L6 = 6.23103


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L7 = 6.39577


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L8 = 6.54242


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L9 = 6.65381


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L10 = 6.72797


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L11 = 6.7979


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L12 = 6.8616


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L13 = 6.90704


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L14 = 6.94762


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L15 = 6.98575


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L16 = 7.01942


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L17 = 7.04536


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L18 = 7.07005


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L19 = 7.09364


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L20 = 7.1136


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L21 = 7.13155


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L22 = 7.14804


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L23 = 7.16441


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L24 = 7.17812


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L25 = 7.19025


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L26 = 7.2019


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L27 = 7.21265


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L28 = 7.22269


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L29 = 7.23251


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


minimum points.

L30 = 7.24162


. . . . . .

Analogies

The 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 better andbetter approximations

The Area Problem (Ch. 5)

I Want the area of a curvedregion

I Only know the area ofpolygons

I Approximate region withpolygons



. . . . . .

Analogies


I Want the slope of a curve

I Only know the slope oflines



The Area Problem (Ch. 5)

I Want the area of a curvedregion





. . . . . .

Analogies


I Want the slope of a curve

I Only know the slope oflines



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





. . . . . .

Analogies


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



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





. . . . . .

Outline









. . . . . .

Distances

Just like area = length× width, we have

distance = rate× time.

So here is another use for Riemann sums.


. . . . . .

Application: Dead Reckoning


. . . . . .

Computing position by Dead Reckoning

Example

A sailing ship is cruising back and forth along a channel (in a straightline). At noon the ship’s position and velocity are recorded, but shortlythereafter a storm blows in and position is impossible to measure. Thevelocity continues to be recorded at thirty-minute intervals.

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

Estimate the ship’s position at 4:00pm.


. . . . . .

Solution

SolutionWe estimate that the speed of 4 knots (nautical miles per hour) ismaintained from 12:00 until 12:30. So over this time interval the shiptravels (

4 nmihr

)(12hr)

= 2nmi

We can continue for each additional 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 15.5nmi east of its original position.


. . . . . .

Analysis

I This method of measuring position by recording velocity wasnecessary until global-positioning satellite technology becamewidespread

I If we had velocity estimates at finer intervals, we’d get betterestimates.

I If we had velocity at every instant, a limit would tell us our exactposition relative to the last time we measured it.


. . . . . .

Other uses of Riemann sums

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


. . . . . .

Outline









. . . . . .


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


. . . . . .

Notation/Terminology

∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci)∆x

I∫

— integral sign (swoopy S)

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

I dx — ??? (a parenthesis? an infinitesimal? a variable?)I The process of computing an integral is called integration orquadrature


. . . . . .


∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci)∆x

I∫


I f(x) — integrand

I a and b — limits of integration (a is the lower limit and b theupper limit)



. . . . . .


∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci)∆x

I∫





. . . . . .


∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci)∆x

I∫



I dx — ??? (a parenthesis? an infinitesimal? a variable?)

I The process of computing an integral is called integration orquadrature


. . . . . .


∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci)∆x

I∫





. . . . . .

The limit can be simplified

TheoremIf f is continuous on [a,b] or if f has only finitely many jumpdiscontinuities, then f is integrable on [a,b]; that is, the definite integral∫ b

af(x) dx exists.

So we can find the integral by computing the limit of any sequence ofRiemann sums that we like,


. . . . . .

Example: Integral of x

Example

Find∫ 3

0x dx

Solution

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

3in.

For each i, take xi to represent the function 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

)

= limn→∞

9n2

n∑i=1

i = limn→∞

9n2

· n(n+ 1)2

=92· 1


. . . . . .


Example

Find∫ 3

0x dx

Solution


3in.

For each i, take xi to represent the function on the ith interval.

So∫ 3

0x dx = lim

n→∞Rn = lim

n→∞

n∑i=1


n∑i=1

(3in

) (3n

)

= limn→∞

9n2

n∑i=1

i = limn→∞

9n2

· n(n+ 1)2

=92· 1


. . . . . .


Example

Find∫ 3

0x dx

Solution


3in.


0x dx = lim

n→∞Rn

= limn→∞

n∑i=1


n∑i=1

(3in

) (3n

)

= limn→∞

9n2

n∑i=1

i = limn→∞

9n2

· n(n+ 1)2

=92· 1


. . . . . .


Example

Find∫ 3

0x dx

Solution


3in.


0x dx = lim

n→∞Rn = lim

n→∞

n∑i=1

f(xi)∆x

= limn→∞

n∑i=1

(3in

) (3n

)

= limn→∞

9n2

n∑i=1

i = limn→∞

9n2

· n(n+ 1)2

=92· 1


. . . . . .


Example

Find∫ 3

0x dx

Solution


3in.


0x dx = lim

n→∞Rn = lim

n→∞

n∑i=1


n∑i=1

(3in

) (3n

)

= limn→∞

9n2

n∑i=1

i = limn→∞

9n2

· n(n+ 1)2

=92· 1


. . . . . .


Example

Find∫ 3

0x dx

Solution


3in.


0x dx = lim

n→∞Rn = lim

n→∞

n∑i=1


n∑i=1

(3in

) (3n

)

= limn→∞

9n2

n∑i=1

i

= limn→∞

9n2

· n(n+ 1)2

=92· 1


. . . . . .


Example

Find∫ 3

0x dx

Solution


3in.


0x dx = lim

n→∞Rn = lim

n→∞

n∑i=1


n∑i=1

(3in

) (3n

)

= limn→∞

9n2

n∑i=1

i = limn→∞

9n2

· n(n+ 1)2

=92· 1


. . . . . .

Example: Integral of x2

Example

Find∫ 3

0x2 dx

Solution

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

)

= limx→∞

27n3

n∑i=1

i2 = limx→∞

27n3

· n(n+ 1)(2n+ 1)6

=273

= 9


. . . . . .


Example

Find∫ 3

0x2 dx

Solution


3in.

So

∫ 3

0x2 dx = lim

x→∞Rn = lim

x→∞

n∑i=1


n∑i=1

(3in

)2(3n

)

= limx→∞

27n3

n∑i=1

i2 = limx→∞

27n3

· n(n+ 1)(2n+ 1)6

=273

= 9


. . . . . .


Example

Find∫ 3

0x2 dx

Solution


3in. So

∫ 3

0x2 dx = lim

x→∞Rn

= limx→∞

n∑i=1


n∑i=1

(3in

)2(3n

)

= limx→∞

27n3

n∑i=1

i2 = limx→∞

27n3

· n(n+ 1)(2n+ 1)6

=273

= 9


. . . . . .


Example

Find∫ 3

0x2 dx

Solution


3in. So

∫ 3

0x2 dx = lim

x→∞Rn = lim

x→∞

n∑i=1

f(xi)∆x

= limx→∞

n∑i=1

(3in

)2(3n

)

= limx→∞

27n3

n∑i=1

i2 = limx→∞

27n3

· n(n+ 1)(2n+ 1)6

=273

= 9


. . . . . .


Example

Find∫ 3

0x2 dx

Solution


3in. So

∫ 3

0x2 dx = lim

x→∞Rn = lim

x→∞

n∑i=1


n∑i=1

(3in

)2(3n

)

= limx→∞

27n3

n∑i=1

i2 = limx→∞

27n3

· n(n+ 1)(2n+ 1)6

=273

= 9


. . . . . .


Example

Find∫ 3

0x2 dx

Solution


3in. So

∫ 3

0x2 dx = lim

x→∞Rn = lim

x→∞

n∑i=1


n∑i=1

(3in

)2(3n

)

= limx→∞

27n3

n∑i=1

i2

= limx→∞

27n3

· n(n+ 1)(2n+ 1)6

=273

= 9


. . . . . .


Example

Find∫ 3

0x2 dx

Solution


3in. So

∫ 3

0x2 dx = lim

x→∞Rn = lim

x→∞

n∑i=1


n∑i=1

(3in

)2(3n

)

= limx→∞

27n3

n∑i=1

i2 = limx→∞

27n3

· n(n+ 1)(2n+ 1)6

=273

= 9


. . . . . .


Example

Find∫ 3

0x3 dx

Solution

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


. . . . . .

Outline









. . . . . .


Example

Estimate∫ 1

0

41+ x2

dx using M4.

Solution

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

12, x3 =

34, x4 = 1.

So c1 =18, c2 =

38, c3 =

58, c4 =

78.

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


. . . . . .


Example

Estimate∫ 1

0

41+ x2

dx using M4.

Solution

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

12, x3 =

34, x4 = 1.

So c1 =18, c2 =

38, c3 =

58, c4 =

78.

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


. . . . . .


Example

Estimate∫ 1

0

41+ x2

dx using M4.

Solution

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

12, x3 =

34, x4 = 1.

So c1 =18, c2 =

38, c3 =

58, c4 =

78.

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


. . . . . .

Estimating the Definite Integral (Continued)

Example

Estimate∫ 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

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


. . . . . .

Outline









. . . . . .


Theorem (Additive Properties of the Integral)

Let f and g be integrable functions 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.


. . . . . .

Proofs

Proofs.

I When integrating a constant function 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 integral of asum is the sum of the integrals.

I Ditto for constant multiplesI Ditto for differences


. . . . . .

Example

Find∫ 3

0

(x3 − 4.5x2 + 5.5x+ 1

)dx

Solution

∫ 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 function we were estimating the integral of before)


. . . . . .

Theorem of the Day

Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


midpoints.

M15 = 7.49968


. . . . . .

More Properties of the Integral

Conventions: ∫ a

bf(x)dx = −

∫ b

af(x)dx

∫ a

af(x)dx = 0

This allows us to have

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c

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


. . . . . .

Example

Suppose f and g are functions with

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.


. . . . . .

SolutionWe 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


. . . . . .

Outline









. . . . . .


TheoremLet f and g be integrable functions 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)


. . . . . .


TheoremLet f and g be integrable functions 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)


. . . . . .

The integral of a nonnegative function is nonnegative

Proof.If f(x) ≥ 0 for all x in [a,b], thenfor any number of divisions nand choice of sample points{ci}:

Sn =n∑

i=1

f(ci)︸︷︷︸≥0

∆x ≥n∑

i=1

0·∆x = 0

Since Sn ≥ 0 for all n, the limitof {Sn} is nonnegative, too:∫ b

af(x)dx = lim

n→∞Sn︸︷︷︸≥0

≥ 0.. x.......


. . . . . .

The definite integral is “increasing"

Proof.

Let h(x) = f(x)− g(x). Iff(x) ≥ g(x) for all x in [a,b],then h(x) ≥ 0 for all x in [a,b].So by the previous property∫ 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

So∫ b

af(x)dx ≥

∫ b

ag(x)dx.


. . . . . .

Bounding the integral using bounds of the function

Proof.If m ≤ f(x) ≤ M on for all x in[a,b], then by the previousproperty∫ b

amdx ≤

∫ b

af(x)dx ≤

∫ b

aMdx

By Property 8, the integral of aconstant function is the productof the constant and the width ofthe interval. So:

m(b−a) ≤∫ b

af(x)dx ≤ M(b−a)

.. x.

y

.

M

.

f(x)

.

m

..a

..b


. . . . . .

Example

Estimate∫ 2

1

1xdx using the comparison properties.

SolutionSince

12≤ x ≤ 1

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

12· 1 ≤

∫ 2

1

1xdx ≤ 1 · 1


. . . . . .

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.


. . . . . .

Summary

I The definite integral is a limit of Riemann SumsI The definite integral can be estimated with Riemann SumsI The definite integral can be distributed across sums and constantmultiples of functions

I The definite integral can be bounded using bounds for the function


lesson 24: areas, distances, the integral (section 021 slides

Documents