Difference between revisions of "Riemann Sums"

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the area that lies between the lines <math style="vertical-align: 0px">x=0</math>&thinsp; and <math style="vertical-align: -4px">x=1, </math>&thinsp; the <math style="vertical-align: 0px">x </math>-axis <math style="vertical-align: -5px">(y=0) </math>&thinsp;
 
the area that lies between the lines <math style="vertical-align: 0px">x=0</math>&thinsp; and <math style="vertical-align: -4px">x=1, </math>&thinsp; the <math style="vertical-align: 0px">x </math>-axis <math style="vertical-align: -5px">(y=0) </math>&thinsp;
 
and the curve <math style="vertical-align: -4px">y=x^{2}. </math>&thinsp; In order to find this area, we can begin
 
and the curve <math style="vertical-align: -4px">y=x^{2}. </math>&thinsp; In order to find this area, we can begin
with a familiar geometric object - the rectangle. In this case, we
+
with a familiar geometric object: the rectangle. In this case, we
wish to find an area above the interval from 0 to 1. In order to approximate
+
wish to find an area above the interval from <math style="vertical-align: 0px">0</math> to <math style="vertical-align: -1px">1</math>. In order to approximate
this, we can divide the interval into, say 4 shorter intervals of
+
this, we can divide the interval into, say <math style="vertical-align: 0px">4</math> shorter intervals of
 
equal length. Let's call this length <math style="vertical-align: 0px">\Delta x </math>, and since they are
 
equal length. Let's call this length <math style="vertical-align: 0px">\Delta x </math>, and since they are
all the same length, we know that the length of each will be <math style="vertical-align: 0px">1/4. </math>  
+
all the same length, we know that the length of each will be <math style="vertical-align: -5px">1/4. </math>  
  
 
However, in order to define an area, our rectangles require a height
 
However, in order to define an area, our rectangles require a height
 
as well as a width. Most often, calculus teachers will use the function's
 
as well as a width. Most often, calculus teachers will use the function's
 
value at the left or right endpoint for the height of each rectangle,
 
value at the left or right endpoint for the height of each rectangle,
although we could also choose the minimum or maximum value of <math style="vertical-align: 0px">f(x) </math>
+
although we could also choose the minimum or maximum value of <math style="vertical-align: -4px">f(x) </math>
 
on each interval, or perhaps the value at the midpoint of each interval.
 
on each interval, or perhaps the value at the midpoint of each interval.
  
 
Let's approximate this area first using left endpoints. Notice that
 
Let's approximate this area first using left endpoints. Notice that
our leftmost interval is <math style="vertical-align: 0px">[0,1/4], </math> so the height at the left endpoint
+
our leftmost interval is <math style="vertical-align: -5px">[0,1/4], </math> so the height at the left endpoint
is <math style="vertical-align: 0px">f(0)=0. </math> This means the area of our leftmost rectangle is
+
is <math style="vertical-align: -5px">f(0)=0. </math> This means the area of our leftmost rectangle is
  
 
:: <math style="vertical-align: 0px">f(0)\cdot\Delta x\,=\,0\cdot\left({\displaystyle \frac{1}{4}}\right)\,=\,0. </math>  
 
:: <math style="vertical-align: 0px">f(0)\cdot\Delta x\,=\,0\cdot\left({\displaystyle \frac{1}{4}}\right)\,=\,0. </math>  
  
Continuing, the adjacent interval is <math style="vertical-align: 0px">[1/4,1/2]. </math> Now, our left endpoint
+
Continuing, the adjacent interval is <math style="vertical-align: -5px">[1/4,1/2]. </math> Now, our left endpoint
is <math style="vertical-align: 0px">1/4 </math>, and our area is  
+
is <math style="vertical-align: -5px">1/4 </math>, and our area is  
  
 
:: <math style="vertical-align: 0px">{\displaystyle f\left(\frac{1}{4}\right)\cdot\Delta x\,=\,\frac{1}{16}\cdot\frac{1}{4}\,=\,\frac{1}{64}}. </math>
 
:: <math style="vertical-align: 0px">{\displaystyle f\left(\frac{1}{4}\right)\cdot\Delta x\,=\,\frac{1}{16}\cdot\frac{1}{4}\,=\,\frac{1}{64}}. </math>
  
The next interval to the right is <math style="vertical-align: 0px">[1/2,3/4], </math> and as such the left
+
The next interval to the right is <math style="vertical-align: -5px">[1/2,3/4], </math> and as such the left
endpoint is <math style="vertical-align: 0px">1/2, </math> so the area is  
+
endpoint is <math style="vertical-align: -5px">1/2, </math> so the area is  
  
 
:: <math style="vertical-align: 0px">{\displaystyle f\left(\frac{1}{2}\right)\cdot\Delta x\,=\,\frac{1}{4}\cdot\frac{1}{4}\,=\,\frac{1}{16}.} </math>  
 
:: <math style="vertical-align: 0px">{\displaystyle f\left(\frac{1}{2}\right)\cdot\Delta x\,=\,\frac{1}{4}\cdot\frac{1}{4}\,=\,\frac{1}{16}.} </math>  
  
Finally, we have the rightmost rectangle, whose base is the interval <math style="vertical-align: 0px">[3/4,1]. </math> This has <math style="vertical-align: 0px">3/4 </math> as its left endpoint, so its area is  
+
Finally, we have the rightmost rectangle, whose base is the interval <math style="vertical-align: -5px">[3/4,1]. </math> This has <math style="vertical-align: -5px">3/4 </math> as its left endpoint, so its area is  
  
 
:: <math style="vertical-align: 0px">{\displaystyle f\left(\frac{3}{4}\right)\cdot\Delta x\,=\,\frac{9}{16}\cdot\frac{1}{4}\,=\,\frac{9}{64}.} </math>  
 
:: <math style="vertical-align: 0px">{\displaystyle f\left(\frac{3}{4}\right)\cdot\Delta x\,=\,\frac{9}{16}\cdot\frac{1}{4}\,=\,\frac{9}{64}.} </math>  
  
Adding these four rectangles up with sigma <math style="vertical-align: 0px">(\Sigma) </math> notation, we
+
Adding these four rectangles up with sigma <math style="vertical-align: -5px">(\Sigma) </math> notation, we
 
can approximate the area under the curve as
 
can approximate the area under the curve as
  
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Of course, we could also use right endpoints. In this case, we would
 
Of course, we could also use right endpoints. In this case, we would
use the endpoints <math style="vertical-align: 0px">1/4,\,1/2,\,3/4 </math> and 1 for each interval from
+
use the endpoints <math style="vertical-align: -5px">1/4,\,1/2,\,3/4 </math> and 1 for each interval from
 
left to right to find
 
left to right to find
  

Revision as of 15:25, 20 September 2015

Approximating Area Under a Curve

Graphically, we can consider a definite integral, such as

to be the area "under the curve", which might be better said as the area that lies between the lines   and   the -axis   and the curve   In order to find this area, we can begin with a familiar geometric object: the rectangle. In this case, we wish to find an area above the interval from to . In order to approximate this, we can divide the interval into, say shorter intervals of equal length. Let's call this length , and since they are all the same length, we know that the length of each will be

However, in order to define an area, our rectangles require a height as well as a width. Most often, calculus teachers will use the function's value at the left or right endpoint for the height of each rectangle, although we could also choose the minimum or maximum value of on each interval, or perhaps the value at the midpoint of each interval.

Let's approximate this area first using left endpoints. Notice that our leftmost interval is so the height at the left endpoint is This means the area of our leftmost rectangle is

Continuing, the adjacent interval is Now, our left endpoint is , and our area is

The next interval to the right is and as such the left endpoint is so the area is

Finally, we have the rightmost rectangle, whose base is the interval This has as its left endpoint, so its area is

Adding these four rectangles up with sigma notation, we can approximate the area under the curve as

Of course, we could also use right endpoints. In this case, we would use the endpoints and 1 for each interval from left to right to find

Note that in this case, one is an overestimate and one is an underestimate.

Additional Examples with Fixed Numbers of Rectangles

Example 1. Approximate the area under the curve of from -1 to 3 using rectangles and left endpoints.

Solution. Note that our -values range from to , so our length is actually Thus each rectangle will have a base of

This is our first step. This means our intervals from left to right are and Choosing left endpoints, we have

Here is where the idea of ``area under the curve becomes clearer. We actually have a signed area, where area below the -axis is negative, while area above the -axis is positive.

Example 2. Approximate the area under the curve of from to using rectangles and midpoints.

Solution. Here, our -values range from to so

As a result, our intervals from left to right are and More importantly, our midpoints occur at and respectively; this is where we will evaluate the height of each rectangle. Thus

Here is where the idea of ``area under the curve becomes clearer. We actually have a signed area, where area below the -axis is negative, while area above the -axis is positive.

Defining the Integral as a Limit

Although associating the area under the curve with four rectangles gives us a really rough approximation, there's no reason we can't continue to divide (partition) the interval into smaller pieces, and then get closer to the actual area. The graphic on the right shows precisely the idea. We can keep making the base of each rectangle, or smaller and smaller, and we'll get a better approximation. More importantly, we can continue this idea as a limit, leading to the following definition.

Definition. We define the Definite Integral of on written

to be

For an introductory course, we usually have so each rectangle has exactly the same base.

Using the Definition to Evaluate a Definite Integral

Frequently, students will be asked questions such as: Using the definition of the definite integral, find the area under the curve of the function on the interval using right endpoints.

Rather than using ``easier rules, such as the power rule and the Fundamental Theorem of Calculus, this requires us to use the the definition just listed. The first step is to set up our sum. We have and is just an arbitrary natural (or counting) number. This tells us that

For a given , our leftmost interval would start at and be of length This describes the interval On the other hand, our next interval would start where the leftmost stopped, or , and it's length would also be This is the interval

If we call the leftmost interval then we would have Similarly, our second interval would be If we were to continue this rule, we would have that for any we could write This allows us to determine where to choose our height for each interval. Since we are asked to use right endpoints, we would want for for and finally for

This allows us to build the sum. For an arbitrary we would have

Using this result, we now have

Now, we have several important sums explained on another page. These are the sum of the first numbers, the sum of the first squares, and the sum of the first cubes:

Moreover, we have some basic rules for summation. These rules that make sense in simpler notation, such as

or

work the same way in Sigma notation, meaning

and

Before worrying about the limit as , when we write both and the in the denominator are just constants, like the in As a result we have

where we applied the rule for the first squares. Finally, we can look at this as being approximately for large so the limit as is Thus

A Left Point of View

What would change if we approached the above integral through left endpoints, instead of right? We would only be changing our value for For example, the leftmost interval is so our left endpoint is On the other hand, our second interval is so our left endpoint is For the interval our left endpoint is Let's apply the same process to this value. We have

From here, we use the special sums again. This means that

Thus our choice of endpoints makes no difference in the resulting value.

A More Advanced Example

For most Riemann Sum problems in an integral calculus class, will always be the same width, and we will need to use the special sums to evaluate the limit. However, what can we do if we wish to determine the value of

Using rectangles of the same width as shown in the earlier animation would result in a very messy sum which contains a lot of square roots! This makes finding the limit nearly impossible. Instead, we could consider the inverse function to the square root, which is squaring. Instead of choosing let's consider and let

For example, while In particular, since we indexed the leftmost point as this means that

Each will be a different width, but either endpoint would be a square, so taking will not leave a square root in our sum.

Now we have all the pieces. Let's use right endpoints for the height of each rectangle, so Then

You will NEVER see something like this in a first year calculus class, but it is just a reminder that the definition includes the indexed for a reason!