Summation NotationSummation Notation• Also called Also called sigma notationsigma notation

(sigma is a Greek letter (sigma is a Greek letter ΣΣ meaning “sum”) meaning “sum”)The series 2+4+6+8+10 can be written as:The series 2+4+6+8+10 can be written as:

i is called the i is called the index of summationindex of summation Sometimes you will see an n or k here instead Sometimes you will see an n or k here instead

of i.of i.The notation is read:The notation is read:

““the sum from i=1 to 5 of 2i”the sum from i=1 to 5 of 2i”

5

1

2ii goes from 1 i goes from 1

to 5.to 5.

Summation Notation for an Summation Notation for an Infinite SeriesInfinite Series

• Summation notation for the infinite series:Summation notation for the infinite series:

2+4+6+8+10+… would be written as:2+4+6+8+10+… would be written as:

Because the series is infinite, you must use i Because the series is infinite, you must use i from 1 to infinity (from 1 to infinity (∞) instead of stopping at ∞) instead of stopping at

the 5the 5thth term like before. term like before.

1

2i

Examples: Write each series in Examples: Write each series in summation notation.summation notation.

a. 4+8+12+…+100a. 4+8+12+…+100• Notice the series can Notice the series can

be written as:be written as:

4(1)+4(2)+4(3)+…+4(25)4(1)+4(2)+4(3)+…+4(25)

Or 4(i) where i goes Or 4(i) where i goes from 1 to 25.from 1 to 25.

• Notice the series Notice the series can be written as:can be written as:

25

1

4i

...5

4

4

3

3

2

2

1 . b

...14

4

13

3

12

2

11

1

. to1 from goes where1

Or,

ii

i

1 1i

i

ExampleExample: Find the sum of the : Find the sum of the series.series.

• k goes from 5 to 10.k goes from 5 to 10.

• (5(522+1)+(6+1)+(622+1)+(7+1)+(722+1)+(8+1)+(822+1)+(9+1)+(922+1)+(10+1)+(1022+1)+1)

= 26+37+50+65+82+101= 26+37+50+65+82+101

= = 361361

10

5

2 1k

You try some. Find the Sum.

5

1

(2 1)i

i

5

1

( )i

c

5

2

( 1)( 3)k

k k

4

20

1

1k k 4

2 3

1

[( 1) ( 1) ]i

i i

Estimating with Finite Sums

Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2002

Greenfield Village, Michigan

time

velocity

After 4 seconds, the object has gone 12 feet.

Consider an object moving at a constant rate of 3 ft/sec.

Since rate . time = distance:

If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.

ft3 4 sec 12 ft

sec

3t d

If the velocity is not constant,we might guess that the distance traveled is still equalto the area under the curve.

(The units work out.)

211

8V t Example:

We could estimate the area under the curve by drawing rectangles touching at their left corners.

This is called the Left-hand Rectangular Approximation Method (LRAM).

1 11

8

11

2

12

8t v

10

1 11

8

2 11

2

3 12

8Approximate area: 1 1 1 3

1 1 1 2 5 5.758 2 8 4

We could also use a Right-hand Rectangular Approximation Method (RRAM).

11

8

11

2

12

8

Approximate area: 1 1 1 31 1 2 3 7 7.75

8 2 8 4

3

211

8V t

Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM).

1.031251.28125

1.78125

Approximate area:6.625

2.53125

t v

1.031250.5

1.5 1.28125

2.5 1.78125

3.5 2.53125

In this example there are four subintervals.As the number of subintervals increases, so does the accuracy.

211

8V t

211

8V t

Approximate area:6.65624

t v

1.007810.25

0.75 1.07031

1.25 1.19531

1.382811.75

2.25

2.75

3.25

3.75

1.63281

1.94531

2.32031

2.75781

13.31248 0.5 6.65624

width of subinterval

With 8 subintervals:

The exact answer for thisproblem is .6.6

Circumscribed rectangles are all above the curve:

Inscribed rectangles are all below the curve:

When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.

211

8V t

subinterval

partition

The width of a rectangle is called a subinterval.

The entire interval is called the partition.

1

limn

kn

k

f c x

If we let n = number of subintervals, then

1

limn

kn

k

f c x

Leibnitz introduced a simpler notation for the definite integral:

1

limn b

k ank

f c x f x dx

Note that the very small change in x becomes dx.

b

af x dx

IntegrationSymbol

lower limit of integration

upper limit of integration

integrandvariable of integration

(dummy variable)

It is called a dummy variable because the answer does not depend on the variable chosen.

time

velocity

After 4 seconds, the object has gone 12 feet.

Earlier, we considered an object moving at a constant rate of 3 ft/sec.

Since rate . time = distance: 3t d

If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.

ft3 4 sec 12 ft

sec

If the velocity varies:

11

2v t

Distance:21

4s t t

(C=0 since s=0 at t=0)

After 4 seconds:1

16 44

s

8s

1Area 1 3 4 8

2

The distance is still equal to the area under the curve!

Notice that the area is a trapezoid.

211

8v t What if:

We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example.

It seems reasonable that the distance will equal the area under the curve.

211

8

dsv t

dt

31

24s t t

314 4

24s

26

3s

The area under the curve2

63

We can use anti-derivatives to find the area under a curve!

Let’s look at it another way:

a x

Let area under the

curve from a to x.

(“a” is a constant)

aA x

x h

aA x

Then:

a x aA x A x h A x h

x a aA x h A x h A x

xA x h

aA x h

x x h

min f max f

The area of a rectangle drawn under the curve would be less than the actual area under the curve.

The area of a rectangle drawn above the curve would be more than the actual area under the curve.

short rectangle area under curve tall rectangle

min max a ah f A x h A x h f

h

min max a aA x h A x

f fh

min max a aA x h A x

f fh

As h gets smaller, min f and max f get closer together.

0

lim a a

h

A x h A xf x

h

This is the definition

of derivative!

a

dA x f x

dx

Take the anti-derivative of both sides to find an explicit formula for area.

aA x F x c

aA a F a c

0 F a c

F a c initial value

min min a aA x h A x

f fh

As h gets smaller, min f and max f get closer together.

0

lim a a

h

A x h A xf x

h

a

dA x f x

dx

aA x F x c

aA a F a c

0 F a c

F a c aA x F x F a

(Area under curve from a to x ) = (antiderivative at x minus antiderivative at a.)

0

1

limn

k kP

k

f c x

b

af x dx

F x F a

Area

“The essence of mathematics is not to make simple things complicated, but to make complicated things simple.” -- Gudder

Area from x=0to x=1

Example: 2y x

Find the area under the curve from x=1 to x=2.

2 2

1x dx

23

1

1

3x

31 12 1

3 3

8 1

3 3

7

3

Area from x=0to x=2

Area under the curve from x=1 to x=2.

Integrals such as are called definite integrals

because we can find a definite value for the answer.

4 2

1x dx

4 2

1x dx

43

1

1

3x C

3 31 14 1

3 3C C

64 1

3 3C C

63

3 21

The constant always cancels when finding a definite integral, so we leave it out!

Integrals such as are called indefinite integrals

because we can not find a definite value for the answer.

2x dx

2x dx31

3x C

When finding indefinite integrals, we always include the “plus C”.

Definite Integration and Areas

0 1

23 2 xy

It can be used to find an area bounded, in part, by a curve

e.g.

1

0

2 23 dxx gives the area shaded on the graph

The limits of integration . . .

Definite integration results in a value.

Areas

Definite Integration and Areas

. . . give the boundaries of the area.

The limits of integration . . .

0 1

23 2 xy

It can be used to find an area bounded, in part, by a curve

Definite integration results in a value.

Areas

x = 0 is the lower limit( the left hand

boundary )x = 1 is the upper limit(the right hand

boundary )

dxx 23 2

0

1

e.g.

gives the area shaded on the graph

Definite Integration and Areas

. . . give the boundaries of the area.

The limits of integration . . .

0 1

23 2 xy

It can be used to find an area bounded, in part, by a curve

Definite integration results in a value.

Areas

x = 0 is the lower limit( the left hand

boundary )x = 1 is the upper limit(the right hand

boundary )

dxx 23 2

0

1

e.g.

gives the area shaded on the graph

Definite Integration and Areas

0 1

23 2 xy

Finding an area

the shaded area equals 3

The units are usually unknown in this type of question

1

0

2 23 dxxSince

31

0

xx 23

Definite Integration and Areas

“Thus mathematics may be defined as the subject in which we never know what we are talking about, not whether what we are saying is true.” -- Russell

3

2

2xdx20

12

dx3

2

1

3x dx3

1

(2 3)x dx5

3 2

2

( )x x dx1

0

sin d 1

33

2dr

r

22

1

2 3x dx 2 3

0

23

xx dx