Area Under the CurveArea Under the Curve We want to approximate the area between a We want to approximate the area between a

curve (curve (yy==xx22+1) and the +1) and the xx-axis from -axis from xx=0 to =0 to xx=7=7

We will use rectangles to do this.We will use rectangles to do this.

One way will be to choose rectangles whose One way will be to choose rectangles whose heights are taken from the heights are taken from the x-x-coordinate of the coordinate of the right side of the rectangle (Right Sum)right side of the rectangle (Right Sum)

We will let We will let nn be the number of rectangles we use be the number of rectangles we use to approximate the areato approximate the area

Right Sum n=5Right Sum n=5

x

y

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-5

0

5

10

15

20

25

30

35

40

45

50

Area = 157.92

Right Sum n=10Right Sum n=10

x

y

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-5

0

5

10

15

20

25

30

35

40

45

50

Area = 139.055

Right Sum n=20Right Sum n=20

x

y

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-5

0

5

10

15

20

25

30

35

40

45

50

Area = 130.0513

Right Sum n=50Right Sum n=50

x

y

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-5

0

5

10

15

20

25

30

35

40

45

50

Area = 124.7862

Right Sum n=100Right Sum n=100

x

y

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-5

0

5

10

15

20

25

30

35

40

45

50

Area = 123.054

What do you notice?What do you notice?

What will happen as you add more rectangles for a Right What will happen as you add more rectangles for a Right Sum?Sum?

Will this happen for any function? Why or why not?Will this happen for any function? Why or why not?

How many rectangles do we need to get the actual How many rectangles do we need to get the actual area?area?

Can you think of another way to approximate the area?Can you think of another way to approximate the area?

Another way:Another way:

What if we were to use rectangles What if we were to use rectangles whose heights were formed from the whose heights were formed from the xx-coordinate of the Left side of the -coordinate of the Left side of the interval?interval?

We will call these Left SumsWe will call these Left Sums As we go through the Left Sums, As we go through the Left Sums,

what do you notice about the areas?what do you notice about the areas?

Left Sum n=5Left Sum n=5

x

y

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-5

0

5

10

15

20

25

30

35

40

45

50

Area = 89.32

Left sum n=10Left sum n=10

x

y

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-5

0

5

10

15

20

25

30

35

40

45

50

Area = 104.755

Left Sum n=20Left Sum n=20

x

y

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-5

0

5

10

15

20

25

30

35

40

45

50

Area = 112.9013

Left Sum n=50Left Sum n=50

x

y

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-5

0

5

10

15

20

25

30

35

40

45

50

Area = 117.9262

Left Sum n=100Left Sum n=100

x

y

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-5

0

5

10

15

20

25

30

35

40

45

50

Area = 119.624

Let’s look at the areas again:Let’s look at the areas again:

Right SumsRight Sums Left SumsLeft Sums

n=5n=5 157.92157.92 89.3289.32

n=10n=10 139.055139.055 104.755104.755

n=20n=20 130.0513130.0513 112.9013112.9013

n=50n=50 124.7862124.7862 117.9262117.9262

n=100n=100 123.054123.054 119.624119.624

n=1000n=1000 121.5049121.5049 121.1619121.1619

n=5000n=5000 121.3676121.3676 121.299121.299

n=30000n=30000 121.3391121.3391 121.3276121.3276

Another ExampleAnother Example

While it is easiest for computational reasons to While it is easiest for computational reasons to look at Left Sums and Right Sums, theoretically look at Left Sums and Right Sums, theoretically it is necessary to look at it is necessary to look at Upper SumsUpper Sums (where (where each rectangle circumscribes the function) and each rectangle circumscribes the function) and Lower SumsLower Sums (where each rectangle is (where each rectangle is inscribed in the function). Recall that in the inscribed in the function). Recall that in the example seen so far, the Right Sums were also example seen so far, the Right Sums were also Upper Sums and the Left Sums were Lower Upper Sums and the Left Sums were Lower Sums. Under what circumstances would this Sums. Under what circumstances would this not be true? Is it possible for some Upper not be true? Is it possible for some Upper Sums to be left sums and others to be right Sums to be left sums and others to be right sums?sums?

x

y

-3 -2 -1 0 1 2 3

5

10

Area = 31.8292

Here is an example of an Upper Sum. Notice that the rectangles are all formed by choosing the height from the highest point on the graph that the rectangle hits.

x

y

-3 -2 -1 0 1 2 3

5

10

Area = 17.8125

Here is a Lower Sum with 8 rectangles. What do you think will happen if a rectangle starts at -½ and ends at ½? What would the height be for a Lower Sum?

Finding Exact AreasFinding Exact AreasRiemann SumsRiemann Sums

It turns out, as you could see from the table, that if you use It turns out, as you could see from the table, that if you use enough rectangles, the Left Sum will be very close to the enough rectangles, the Left Sum will be very close to the Right Sum. If you use an infinite amount of rectangles, the Right Sum. If you use an infinite amount of rectangles, the Left Sum and Right Sum will be equal, and they will equal the Left Sum and Right Sum will be equal, and they will equal the exact areaexact area..

Theoretically, there are a great many ways to design Theoretically, there are a great many ways to design rectangles that will lead to the area. Any of these will be a rectangles that will lead to the area. Any of these will be a type of Riemann Sum. So far we have seen Left, Right, type of Riemann Sum. So far we have seen Left, Right, Upper, and Lower Sums; these are all types of Riemann Upper, and Lower Sums; these are all types of Riemann Sums. There are lots of others.Sums. There are lots of others.

Since Since anyany Riemann Sum will eventually lead to the exact area, Riemann Sum will eventually lead to the exact area, we will use the Right Sum. The Right Sum is computationally we will use the Right Sum. The Right Sum is computationally easiest to use.easiest to use.

Finding Exact AreasFinding Exact AreasRiemann SumsRiemann Sums

Back toBack to If we use 8 If we use 8

rectangles, how wide rectangles, how wide is each one?is each one?

What about if we What about if we used 12 rectangles?used 12 rectangles?

How about How about nn rectangles?rectangles?

In general, if we start In general, if we start at at x=ax=a and stop at and stop at x=bx=b, how wide would , how wide would each rectangle be?each rectangle be?

x

y

0 1 2 3 4 5 6 7

10

20

30

40

50

y x 2 1

Exact AreaExact Area

In general, the width of each In general, the width of each rectangle will berectangle will be

We call the width of each We call the width of each rectangle rectangle

What would the height of each What would the height of each rectangle be? Let’s look at the rectangle be? Let’s look at the graph again, and then see if we graph again, and then see if we can generalize… can generalize…

b a

n

x

As you can see, the heights of each rectangle are found by getting the y-value at the right side of each rectangle. If xi represents the x-coordinate of the right side of the ith rectangle, then the height of the ith rectangle is f(xi ). For our example, each xi

is found by adding to the previous right side. Since we start at x=0, the first right side is the next one is . Without doing the addition, what would be the coordinate of the right side of the 5th rectangle? Do you see that you merely multiply 5 times ? How would this generalize?

78787

878

148

148

x

y

0

0

5

10

15

20

25

30

35

40

45

50

7878

If the area we are interested starts at If the area we are interested starts at x=ax=a, and the width is , and the width is

then the then the xx-coordinate of the-coordinate of the iith th rectangle rectangle xxii will be will be

Example: If I was finding the area under a curve on the Example: If I was finding the area under a curve on the

interval [3, 7] and I was using 100 rectangles, the interval [3, 7] and I was using 100 rectangles, the xx--

coordinate of the 70coordinate of the 70thth rectangle would be rectangle would be

Find Find xx2525 for an area on the interval [2, 8] if we use 120 for an area on the interval [2, 8] if we use 120

rectanglesrectangles

Did you get 3.25?Did you get 3.25?

b an

xiaxn

abiax ii

or

545or

25

703or

100

37703

Area of the Area of the i i thth rectangle rectangle

Since area of a rectangle is height times width, and the Since area of a rectangle is height times width, and the

height is just the value of the function at height is just the value of the function at xxi i , we get, we get

Given on the interval [5, 7] with 20 Given on the interval [5, 7] with 20

rectangles, find the area of the 14rectangles, find the area of the 14thth rectangle. rectangle.

Did you get Did you get

n

ab

n

abiafxxiafxxfA ii

)()(

13)( 2 xxxf

? 116.6or 250

1529

Putting it all togetherPutting it all together

Now we want to put it all together. We want to add up all Now we want to put it all together. We want to add up all nn rectangles to give an approximation for the area. This is rectangles to give an approximation for the area. This is the formula we use:the formula we use:

Let’s go back to our first example: on [0, 7] and Let’s go back to our first example: on [0, 7] and let’s use 100 rectangles. We get let’s use 100 rectangles. We get

which is what we had before on our table.which is what we had before on our table.

n

ab

n

abiafxxf

n

i

n

ii

or )(11

12 xy

123.05405 10016

201101100

10000

49

100

7

110000

49

100

7

100

07 1

100

070

100

1

100

1

2100

1

2

iii

ii

The exact areaThe exact area

To find the exact area, all we need to do is look at an To find the exact area, all we need to do is look at an infinite number of rectangles. Believe it or not, this is infinite number of rectangles. Believe it or not, this is actually easier than what we just did. The formula actually easier than what we just did. The formula becomesbecomes

The previous example becomesThe previous example becomes

Remember that the sum of a constant is the constant times Remember that the sum of a constant is the constant times nn. .

n

ab

n

abiaf

nxxf

n

n

i

n

ii

limor )(

lim

11

33.12176

23431

7

6

)12)(1(343lim

1497lim07

107

0lim

3

11

22

1

2

nn

nnn

nn

innnnn

in

n

i

n

i

n

i

Another exampleAnother example

Find the area underFind the area under on theon the interval [2, 5]interval [2, 5]

We have , andWe have , and

So the area is So the area is

432 2 xxy

nnx

325

61518

43

233

22)(2

22

n

i

n

iin

in

xf i

in

xi3

2

5.58182

45

6

108

18

2

)1(45

6

)12()1(54

lim

184554i

lim

3

61518

lim

23

2

n

1i3

2n

1i2

2

nn

nn

n

nnn

nn

nn

i

nnnn

i

n

i

n

A few last commentsA few last comments Don’t Panic – an easier way is coming soonDon’t Panic – an easier way is coming soon There are many, many applications of what we just did…also There are many, many applications of what we just did…also

coming sooncoming soon What do you think would change if we tried to find the exact area What do you think would change if we tried to find the exact area

using a Left Sum? Why is using a Left Sum more complicated?using a Left Sum? Why is using a Left Sum more complicated?

Another, very common method used to approximate the area Another, very common method used to approximate the area under a curve is called a Midpoint Sum. Without any other under a curve is called a Midpoint Sum. Without any other information, what do you think that might be? information, what do you think that might be?

To To approximateapproximate the area under a curve, we usually just the area under a curve, we usually just use a few rectangles, and which method we use use a few rectangles, and which method we use depends on what the graph looks like. To get the exact depends on what the graph looks like. To get the exact area we use Right Sums, but we are limited to only area we use Right Sums, but we are limited to only finding the area under polynomial of degree 3 or less finding the area under polynomial of degree 3 or less (Why?)(Why?)

A final comment or threeA final comment or three

If you are approximating the area under a curve and If you are approximating the area under a curve and

you are using only a few rectangles it is you are using only a few rectangles it is MUCHMUCH easier to find the heights and areas by hand (or easier to find the heights and areas by hand (or using a table in your calculator) than to use using a table in your calculator) than to use formulas.formulas.

The width of each rectangle will not always be the The width of each rectangle will not always be the same…think about how you approach that situation.same…think about how you approach that situation.

On an AP test you will need to do Left, Right, On an AP test you will need to do Left, Right, Upper, Lower, and/or Midpoint Sums for Upper, Lower, and/or Midpoint Sums for smallsmall numbers of rectangles.numbers of rectangles.