Difference between revisions of "Riemann Sums"

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== Approximating Area Under a Curve ==
 
== Approximating Area Under a Curve ==
 
+
[[File:TallRiem.png|right|300px]]
 
Graphically, we can consider a definite integral, such as  
 
Graphically, we can consider a definite integral, such as  
  
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to be the area "under the curve", which might be better said as
 
to be the area "under the curve", which might be better said as
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 line <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
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:: <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: -5px">(\Sigma) </math> notation, we
+
Adding these four rectangles up with sigma <math style="vertical-align: -5px">(\Sigma) </math>&thinsp; 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: -5px">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 <math style="vertical-align: -1px">1</math> for the height above each interval from
 
left to right to find
 
left to right to find
  
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== Additional Examples with Fixed Numbers of Rectangles ==
 
== Additional Examples with Fixed Numbers of Rectangles ==
  
Example 1. Approximate the area under the curve of <math style="vertical-align: 0px">f(x)=x^{3}-x </math>
+
'''Example 1.''' Approximate the area under the curve of <math style="vertical-align: -5px">f(x)=x^{3}-x </math>
from -1 to 3 using <math style="vertical-align: 0px">n=4 </math> rectangles and left endpoints.
+
from &thinsp;<math style="vertical-align: -1px">-1</math> to <math style="vertical-align: 0px">3</math>&thinsp; using <math style="vertical-align: -1px">n=4 </math> rectangles and left endpoints.
  
Solution. Note that our <math style="vertical-align: 0px">x </math>-values range from <math style="vertical-align: 0px">-1 </math> to <math style="vertical-align: 0px">3 </math>, so our
+
'''Solution.''' Note that our <math style="vertical-align: 0px">x </math>-values range from <math style="vertical-align: -1px">-1 </math> to <math style="vertical-align: 0px">3 </math>, so our
length is actually <math style="vertical-align: 0px">3-(-1)=4. </math> Thus each rectangle will have a base
+
length is actually <math style="vertical-align: -5px">3-(-1)=4. </math> Thus each rectangle will have a base
 
of  
 
of  
  
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This is our first step. This means our intervals from left to right
 
This is our first step. This means our intervals from left to right
are <math style="vertical-align: 0px">[-1,0],\,[0,1],\,[1,2] </math> and <math style="vertical-align: 0px">[2,3]. </math> Choosing left endpoints,
+
are <math style="vertical-align: -4px">[-1,0],\,[0,1],\,[1,2] </math> and <math style="vertical-align: -4px">[2,3]. </math> Choosing left endpoints,
 
we have
 
we have
  
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\end{array} </math>
 
\end{array} </math>
  
Here is where the idea of ``area under the curve'' becomes clearer.
+
Here is where the idea of "area under the curve" becomes clearer.
 
We actually have a signed area, where area below the <math style="vertical-align: 0px">x </math>-axis is
 
We actually have a signed area, where area below the <math style="vertical-align: 0px">x </math>-axis is
 
negative, while area above the <math style="vertical-align: 0px">x </math>-axis is positive.
 
negative, while area above the <math style="vertical-align: 0px">x </math>-axis is positive.
  
Example 2. Approximate the area under the curve of <math style="vertical-align: 0px">f(x)=x^{3}-x </math>
+
'''Example 2.''' Approximate the area under the curve of <math style="vertical-align: -5px">f(x)=x^{3}-x </math>
from <math style="vertical-align: 0px">-4 </math> to <math style="vertical-align: 0px">4 </math> using <math style="vertical-align: 0px">n=4 </math> rectangles and midpoints.
+
from <math style="vertical-align: -1px">-4 </math> to <math style="vertical-align: -1px">4 </math> using <math style="vertical-align: -1px">n=4 </math> rectangles and midpoints.
  
Solution. Here, our <math style="vertical-align: 0px">x </math>-values range from <math style="vertical-align: 0px">-4 </math> to <math style="vertical-align: 0px">4, </math> so  
+
'''Solution.''' Here, our <math style="vertical-align: 0px">x </math>-values range from <math style="vertical-align: -1px">-4 </math>&thinsp; to <math style="vertical-align: -4px">4, </math> so  
  
 
:: <math style="vertical-align: 0px">\Delta x\,=\,{\displaystyle \frac{b-a}{n}\,=\,\frac{4-(-4)}{4}\,=\,\frac{8}{4}\,=\,2.} </math>  
 
:: <math style="vertical-align: 0px">\Delta x\,=\,{\displaystyle \frac{b-a}{n}\,=\,\frac{4-(-4)}{4}\,=\,\frac{8}{4}\,=\,2.} </math>  
  
As a result, our intervals from left to right are <math style="vertical-align: 0px">[-4,-2],\,[-2,0],\,[0,2] </math>
+
As a result, our intervals from left to right are <math style="vertical-align: -4px">[-4,-2],\,[-2,0],\,[0,2] </math>
and <math style="vertical-align: 0px">[2,4]. </math> More importantly, our midpoints occur at <math style="vertical-align: 0px">-3,\,-1,\,1 </math>
+
and &thinsp;<math style="vertical-align: -4px">[2,4]. </math> More importantly, our midpoints occur at <math style="vertical-align: -4px">-3,\,-1,\,1 </math>
 
and <math style="vertical-align: 0px">3 </math> respectively; this is where we will evaluate the height of
 
and <math style="vertical-align: 0px">3 </math> respectively; this is where we will evaluate the height of
 
each rectangle. Thus
 
each rectangle. Thus
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\\
 
\\
 
  & = & 0.
 
  & = & 0.
\end{array} </math>  
+
\end{array} </math>
  
Here is where the idea of ``area under the curve'' becomes clearer.
+
== Defining the Integral as a Limit ==
We actually have a signed area, where area below the <math style="vertical-align: 0px">x </math>-axis is
 
negative, while area above the <math style="vertical-align: 0px">x </math>-axis is positive.
 
  
== Defining the Integral as a Limit ==
+
[[File:Riemann.gif|right|500px]]
  
 
Although associating the area under the curve with four rectangles
 
Although associating the area under the curve with four rectangles
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then get closer to the actual area. The graphic on the right shows
 
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,
 
precisely the idea. We can keep making the base of each rectangle,
or <math style="vertical-align: 0px">\Delta x, </math> smaller and smaller, and we'll get a better approximation.
+
or <math style="vertical-align: -4px">\Delta x, </math> smaller and smaller, and we'll get a better approximation.
 
More importantly, we can continue this idea as a limit, leading to
 
More importantly, we can continue this idea as a limit, leading to
 
the following definition.  
 
the following definition.  
  
Definition. We define the Definite Integral of <math style="vertical-align: 0px">f(x) </math> on <math style="vertical-align: 0px">[a,b], </math>
+
Definition. We define the '''Definite Integral''' of <math style="vertical-align: -4px">f(x) </math> on <math style="vertical-align: -5px">[a,b], </math>&thinsp;
 
written
 
written
  
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:: <math style="vertical-align: 0px">{\displaystyle \int_{a}^{b}f(x)\, dx\,=\,\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x_{i})\cdot\Delta x_{i}.} </math>
 
:: <math style="vertical-align: 0px">{\displaystyle \int_{a}^{b}f(x)\, dx\,=\,\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x_{i})\cdot\Delta x_{i}.} </math>
  
For an introductory course, we usually have <math style="vertical-align: 0px">\Delta x_{i}=\Delta x={\displaystyle \frac{b-a}{n},} </math>
+
For an introductory course, we usually have <math style="vertical-align: -14px">\Delta x_{i}=\Delta x={\displaystyle \frac{b-a}{n},} </math>&thinsp;
so each rectangle has exactly the same base.  
+
so each rectangle has exactly the same base.
  
 
== Using the Definition to Evaluate a Definite Integral ==
 
== Using the Definition to Evaluate a Definite Integral ==
  
 
Frequently, students will be asked questions such as: Using the definition
 
Frequently, students will be asked questions such as: Using the definition
of the definite integral, find the area under the curve of the function <math style="vertical-align: 0px">f(x)=x^{2} </math> on the interval <math style="vertical-align: 0px">[0,3] </math> using right endpoints.  
+
of the definite integral, find the area under the curve of the function <math style="vertical-align: -4px">f(x)=x^{2} </math> on the interval <math style="vertical-align: -5px">[0,3] </math> using right endpoints.  
  
Rather than using ``easier'' rules, such as the power rule and the
+
Rather than using "easier" rules, such as the power rule and the
 
Fundamental Theorem of Calculus, this requires us to use the the definition
 
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 <math style="vertical-align: 0px">a=0,\, b=3 </math>
+
just listed. The first step is to set up our sum. We have <math style="vertical-align: -4px">a=0,\, b=3 </math>&thinsp;
 
and <math style="vertical-align: 0px">n </math> is just an arbitrary natural (or counting) number. This tells
 
and <math style="vertical-align: 0px">n </math> is just an arbitrary natural (or counting) number. This tells
 
us that  
 
us that  
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:: <math style="vertical-align: 0px">\Delta x\,=\,{\displaystyle \frac{b-a}{n}\,=\,\frac{3-0}{n}\,=\,\frac{3}{n}}. </math>  
 
:: <math style="vertical-align: 0px">\Delta x\,=\,{\displaystyle \frac{b-a}{n}\,=\,\frac{3-0}{n}\,=\,\frac{3}{n}}. </math>  
  
For a given <math style="vertical-align: 0px">n </math>, our leftmost interval would start at <math style="vertical-align: 0px">0, </math> and be
+
For a given <math style="vertical-align: 0px">n </math>, our leftmost interval would start at <math style="vertical-align: -5px">0, </math> and be
of length <math style="vertical-align: 0px">\Delta x=3/n. </math> This describes the interval <math style="vertical-align: 0px">[0,3/n]. </math>
+
of length <math style="vertical-align: -5px">\Delta x=3/n. </math> This describes the interval <math style="vertical-align: -5px">[0,3/n]. </math>
 
On the other hand, our next interval would start where the leftmost
 
On the other hand, our next interval would start where the leftmost
stopped, or <math style="vertical-align: 0px">3/n </math>, and it's length would also be <math style="vertical-align: 0px">\Delta x=3/n. </math>
+
stopped, or <math style="vertical-align: -5px">3/n </math>, and it's length would also be <math style="vertical-align: -5px">\Delta x=3/n. </math>
 
This is the interval  
 
This is the interval  
  
 
:: <math style="vertical-align: 0px">{\displaystyle \left[\frac{3}{n},\frac{3}{n}+\frac{3}{n}\right]\,=\,\left[1\cdot\frac{3}{n},2\cdot\frac{3}{n}\right].} </math>  
 
:: <math style="vertical-align: 0px">{\displaystyle \left[\frac{3}{n},\frac{3}{n}+\frac{3}{n}\right]\,=\,\left[1\cdot\frac{3}{n},2\cdot\frac{3}{n}\right].} </math>  
  
If we call the leftmost interval <math style="vertical-align: 0px">I_{1}, </math> then we would have <math style="vertical-align: 0px">I_{1}={\displaystyle \left[0\cdot\frac{3}{n},1\cdot\frac{3}{n}\right]}. </math>
+
If we call the leftmost interval <math style="vertical-align: -4px">I_{1}, </math> then we would have <math style="vertical-align: -15px">I_{1}={\displaystyle \left[0\cdot\frac{3}{n},1\cdot\frac{3}{n}\right]}. </math>
Similarly, our second interval would be <math style="vertical-align: 0px">I_{2}={\displaystyle \left[1\cdot\frac{3}{n},2\cdot\frac{3}{n}\right]}. </math>
+
Similarly, our second interval would be <math style="vertical-align: -15px">I_{2}={\displaystyle \left[1\cdot\frac{3}{n},2\cdot\frac{3}{n}\right]}. </math>
If we were to continue this rule, we would have that for any <math style="vertical-align: 0px">i=1,2,\ldots,n, </math>
+
If we were to continue this rule, we would have that for any <math style="vertical-align: -5px">i=1,2,\ldots,n, </math>
we could write <math style="vertical-align: 0px">I_{i}={\displaystyle \left[(i-1)\cdot\frac{3}{n},i\cdot\frac{3}{n}\right]}. </math>
+
we could write <math style="vertical-align: -15px">I_{i}={\displaystyle \left[(i-1)\cdot\frac{3}{n},i\cdot\frac{3}{n}\right]}. </math>
 
This allows us to determine where to choose our height for each interval.
 
This allows us to determine where to choose our height for each interval.
Since we are asked to use right endpoints, we would want <math style="vertical-align: 0px">f\left({\displaystyle 1\cdot\frac{3}{n}}\right) </math>
+
Since we are asked to use right endpoints, we would want <math style="vertical-align: -15px">f\left({\displaystyle 1\cdot\frac{3}{n}}\right) </math>
for <math style="vertical-align: 0px">I_{1}, </math> <math style="vertical-align: 0px">f\left({\displaystyle 2\cdot\frac{3}{n}}\right) </math> for <math style="vertical-align: 0px">I_{2}, </math> and finally <math style="vertical-align: 0px">f\left({\displaystyle i\cdot\frac{3}{n}}\right) </math>
+
for <math style="vertical-align: -4px">I_{1}, </math> <math style="vertical-align: -15px">f\left({\displaystyle 2\cdot\frac{3}{n}}\right) </math> for <math style="vertical-align: -4px">I_{2}, </math> and finally <math style="vertical-align: -15px">f\left({\displaystyle i\cdot\frac{3}{n}}\right) </math>
for <math style="vertical-align: 0px">I_{i}. </math>
+
for <math style="vertical-align: -4px">I_{i}. </math>
  
This allows us to build the sum. For an arbitrary <math style="vertical-align: 0px">n, </math> we would have  
+
This allows us to build the sum. For an arbitrary <math style="vertical-align: -5px">n, </math> we would have  
  
 
:: <math style="vertical-align: 0px">{\displaystyle \sum_{i=1}^{n}f(x_{i})\cdot\Delta x_{i}\,=\,\sum_{i=1}^{n}f\left(\frac{3i}{n}\right)\cdot\Delta x\,=\,\sum_{i=1}^{n}\frac{9i^{2}}{n^{2}}\cdot\frac{3}{n}\,=\,\sum_{i=1}^{n}\frac{27i^{2}}{n^{3}}.} </math>
 
:: <math style="vertical-align: 0px">{\displaystyle \sum_{i=1}^{n}f(x_{i})\cdot\Delta x_{i}\,=\,\sum_{i=1}^{n}f\left(\frac{3i}{n}\right)\cdot\Delta x\,=\,\sum_{i=1}^{n}\frac{9i^{2}}{n^{2}}\cdot\frac{3}{n}\,=\,\sum_{i=1}^{n}\frac{27i^{2}}{n^{3}}.} </math>
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:: <math style="vertical-align: 0px">{\displaystyle \sum_{i=1}^{n}(a_{i}+b_{i})\,=\, a_{1}+b_{1}+a_{2}+b_{2}\cdots a_{n}+b_{n}\,=\, a_{1}+a_{2}+\cdots+a_{n}+b_{1}+b_{2}+\cdots+b_{n}\,=\,\sum_{i=1}^{n}a_{i}+\sum_{i=1}^{n}b_{i}.\qquad\qquad(\dagger\dagger)} </math>
 
:: <math style="vertical-align: 0px">{\displaystyle \sum_{i=1}^{n}(a_{i}+b_{i})\,=\, a_{1}+b_{1}+a_{2}+b_{2}\cdots a_{n}+b_{n}\,=\, a_{1}+a_{2}+\cdots+a_{n}+b_{1}+b_{2}+\cdots+b_{n}\,=\,\sum_{i=1}^{n}a_{i}+\sum_{i=1}^{n}b_{i}.\qquad\qquad(\dagger\dagger)} </math>
  
Before worrying about the limit as <math style="vertical-align: 0px">n\rightarrow\infty </math>, when we
+
Before worrying about the limit as <math style="vertical-align: -1px">n\rightarrow\infty </math>, when we
write <math style="vertical-align: 0px">{\displaystyle \sum_{i=1}^{n}\frac{27i^{2}}{n^{3}},} </math> both <math style="vertical-align: 0px">27 </math> and the <math style="vertical-align: 0px">n^{3} </math> in the denominator are just constants, like
+
write <math style="vertical-align: -20px">{\displaystyle \sum_{i=1}^{n}\frac{27i^{2}}{n^{3}},} </math> both <math style="vertical-align: 0px">27 </math> and the <math style="vertical-align: 0px">n^{3} </math> in the denominator are just constants, like
 
the <math style="vertical-align: 0px">c </math> in <math style="vertical-align: 0px">(\dagger). </math> As a result we have  
 
the <math style="vertical-align: 0px">c </math> in <math style="vertical-align: 0px">(\dagger). </math> As a result we have  
  
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where we applied the rule for the first <math style="vertical-align: 0px">n </math> squares. Finally, we
 
where we applied the rule for the first <math style="vertical-align: 0px">n </math> squares. Finally, we
can look at this as being approximately <math style="vertical-align: 0px">18n^{3}/2n^{3} </math> for large <math style="vertical-align: 0px">n, </math> so the limit as <math style="vertical-align: 0px">n\rightarrow\infty </math> is <math style="vertical-align: 0px">9. </math> Thus
+
can look at this as being approximately <math style="vertical-align: -5px">18n^{3}/2n^{3} </math>&thinsp; for large <math style="vertical-align: -4px">n, </math> so the limit as <math style="vertical-align: -1px">n\rightarrow\infty </math>&thinsp; is <math style="vertical-align: 0px">9. </math> Thus
  
 
:: <math style="vertical-align: 0px">{\displaystyle \int_{0}^{3}x^{2}\, dx=9.} </math>
 
:: <math style="vertical-align: 0px">{\displaystyle \int_{0}^{3}x^{2}\, dx=9.} </math>
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What would change if we approached the above integral through left
 
What would change if we approached the above integral through left
endpoints, instead of right? We would only be changing our value for <math style="vertical-align: 0px">f(x_{i}). </math> For example, the leftmost interval is <math style="vertical-align: 0px">{\displaystyle \left[0\cdot\frac{3}{n},1\cdot\frac{3}{n}\right],} </math>
+
endpoints, instead of right? We would only be changing our value for <math style="vertical-align: -5px">f(x_{i}). </math> For example, the leftmost interval is&thinsp; <math style="vertical-align: -15px">{\displaystyle \left[0\cdot\frac{3}{n},1\cdot\frac{3}{n}\right],} </math>
 
so our left endpoint is <math style="vertical-align: 0px">0. </math> On the other hand, our second interval
 
so our left endpoint is <math style="vertical-align: 0px">0. </math> On the other hand, our second interval
is <math style="vertical-align: 0px">{\displaystyle \left[1\cdot\frac{3}{n},2\cdot\frac{3}{n}\right],} </math>
+
is&thinsp; <math style="vertical-align: --15px">{\displaystyle \left[1\cdot\frac{3}{n},2\cdot\frac{3}{n}\right],} </math>
so our left endpoint is <math style="vertical-align: 0px">3/n. </math> For the interval <math style="vertical-align: 0px">I_{1}={\displaystyle \left[(i-1)\cdot\frac{3}{n},i\cdot\frac{3}{n}\right],} </math>
+
so our left endpoint is <math style="vertical-align: -5px">3/n. </math> For the interval <math style="vertical-align: -15px">I_{1}={\displaystyle \left[(i-1)\cdot\frac{3}{n},i\cdot\frac{3}{n}\right],} </math>
our left endpoint is <math style="vertical-align: 0px">3(i-1)/n. </math> Let's apply the same process to
+
our left endpoint is <math style="vertical-align: -5px">3(i-1)/n. </math> Let's apply the same process as the last section to  
this value. We have  
+
this value. for height. We then have  
  
 
:: <math style="vertical-align: 0px">\begin{array}{rcl}
 
:: <math style="vertical-align: 0px">\begin{array}{rcl}
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  & = & {\displaystyle \lim_{n\rightarrow\infty}\,\left(\frac{9n(n+1)(2n+1)}{2n^{3}}+\frac{27n(n+1)}{2n^{3}}+\frac{27}{n^{2}}\right)}\\
 
  & = & {\displaystyle \lim_{n\rightarrow\infty}\,\left(\frac{9n(n+1)(2n+1)}{2n^{3}}+\frac{27n(n+1)}{2n^{3}}+\frac{27}{n^{2}}\right)}\\
 
\\
 
\\
  & = & {\displaystyle \lim_{n\rightarrow\infty}\,\left(\frac{18n^{3}}{2n^{3}}+\frac{27n^{2}}{2n^{3}}+\frac{27}{n^{2}}\right)\qquad\qquad(\textrm{for large }n)}\\
+
  & = & {\displaystyle \lim_{n\rightarrow\infty}\,\left(\frac{18n^{3}}{2n^{3}}+\frac{27n^{2}}{2n^{3}}+\frac{27}{n^{2}}\right)\qquad\qquad(\textrm{for~large~}n)}\\
 
\\
 
\\
 
  & = & 9+0+0\\
 
  & = & 9+0+0\\
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Thus our choice of endpoints makes no difference in the resulting
 
Thus our choice of endpoints makes no difference in the resulting
value.  
+
value.
  
 
== A More Advanced Example ==
 
== A More Advanced Example ==

Latest revision as of 21:35, 27 September 2015

Approximating Area Under a Curve

TallRiem.png

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 line   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 for the height above each interval from left to right to find

Note that in this case, one is an overestimate and one is an underestimate. This approximation through the area of rectangles is known as a Riemann sum.

Additional Examples with Fixed Numbers of Rectangles

Example 1. Approximate the area under the curve of from   to   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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x } -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

Defining the Integral as a Limit

Riemann.gif

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 as the last section to this value. for height. We then 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; the rectangles don't need to be all of the same width!