Editing Darboux integral (section)

==Examples==

===A Darboux-integrable function===
Suppose we want to show that the function <math>f(x)=x</math> is Darboux-integrable on the interval <math>[0,1]</math> and determine its value. To do this we partition <math>[0,1]</math> into <math>n</math> equally sized subintervals each of length <math>1/n</math>. We denote a partition of <math>n</math> equally sized subintervals as <math>P_n</math>.

Now since <math>f(x)=x</math> is strictly increasing on <math>[0,1]</math>, the infimum on any particular subinterval is given by its starting point. Likewise the supremum on any particular subinterval is given by its end point. The starting point of the <math>k</math>-th subinterval in <math>P_n</math> is <math>(k-1)/n</math> and the end point is <math>k/n</math>. Thus the lower Darboux sum on a partition <math>P_n</math> is given by

:<math>\begin{align}
L_{f,P_n} &= \sum_{k = 1}^{n} f(x_{k-1})(x_{k} - x_{k-1}) \\
          &= \sum_{k = 1}^{n} \frac{k-1}{n} \cdot \frac{1}{n} \\
          &= \frac{1}{n^2} \sum_{k = 1}^{n} [k-1] \\ 
          &= \frac{1}{n^2}\left[ \frac{(n-1)n}{2} \right] \\
          &= \frac{1}{2} - \frac{1}{2n}
\end{align}</math>

similarly, the upper Darboux sum is given by

:<math>\begin{align}
U_{f,P_n} &= \sum_{k = 1}^{n} f(x_{k})(x_{k} - x_{k-1}) \\
          &= \sum_{k = 1}^{n} \frac{k}{n} \cdot \frac{1}{n} \\
          &= \frac{1}{n^2} \sum_{k = 1}^{n} k \\ 
          &= \frac{1}{n^2}\left[ \frac{(n+1)n}{2} \right] \\
          &= \frac{1}{2} + \frac{1}{2n}
\end{align}</math>

Since
:<math>U_{f,P_n} - L_{f,P_n} = \frac{1}{n}</math>
Thus for given any <math>\varepsilon>0</math>, we have that any partition <math>P_n</math> with <math>n > \frac{1}{\varepsilon}</math> satisfies 
:<math>U_{f,P_n} - L_{f,P_n} < \varepsilon</math>
which shows that <math>f</math> is Darboux integrable. To find the value of the integral note that

:<math>\int_{0}^{1}f(x) \, dx
= \lim_{n \to \infty} U_{f,P_n}
= \lim_{n \to \infty} L_{f,P_n}
= \frac{1}{2}</math>
{{multiple image
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|image1=Riemann Integration and Darboux Upper Sums.gif
|width1=300
|caption1=Darboux upper sums of the function {{math|1=''y'' = ''x''<sup>2</sup>}}
|alt1=Upper Darboux sum example

|image2=Riemann Integration and Darboux Lower Sums.gif
|width2=300
|caption2=Darboux lower sums of the function {{math|1=''y'' = ''x''<sup>2</sup>}}
|alt2=Lower Darboux sum example
}}

===A nonintegrable function===
Suppose we have the [[Dirichlet function]] <math>f:\R \to [0,1]</math> defined as 

:<math>\begin{align}
f(x) &=
 \begin{cases}
 1 & \text{if }x\text{ is rational} \\
 0 & \text{if }x\text{ is irrational}
 \end{cases}
\end{align}</math>

Since the rational and irrational numbers are both [[dense subset]]s of <math>\mathbb{R}</math>, it follows that <math>f</math> takes on the value of 0 and 1 on every subinterval of any partition. Thus for any partition <math>P</math> we have

:<math>\begin{align}
L_{f,P} &=\sum_{k = 1}^{n}(x_{k} - x_{k-1})\inf_{x \in [x_{k-1},x_{k}]}f = 0 \\
U_{f,P} &=\sum_{k = 1}^{n}(x_{k} - x_{k-1}) \sup_{x \in [x_{k-1},x_{k}]}f = 1
\end{align}</math>
from which we can see that the lower and upper Darboux integrals are unequal.