A Riemann ("ree-mahn") sum is a mathematical technique for approximating the area under a function or curve, named after the German mathematician Georg Friedrich Bernhard Riemann. The technique involves estimating the area under a function by splitting it into variably sized rectangles whose areas can easily be found. Riemann sums are important to Calculus since they are the precursors to definite integrals or antiderivatives, which are really just Riemann sums made acurate by narrowing the rectangles to an infinitely small width. Riemann sums are taken relative to the x-axis, and areas between a function below the x-axis and the x-axis are considered positive.

The basis of the Riemann integral. Given a function f:[a,b]->R, a partition P={a=x0<x1<...<xn=b} of the interval [a,b], and points C={ci∈[xi-1,xi] : i=1,...,n}, we wish to approximate (what will eventually become) the Riemann integral ∫ab f(x)dx.

The Riemann sum of f with respect to the above partition {xi} and points {ci} is the sum

I(f;x0,...,xn;c1,...,cn) = ∑i=1n (xi-xi-1)f(ci)
That is, approximate the value of f in each subinterval of the partition by its value at some point in the subinterval (this might make sense e.g. if f is continuous), and replace the area under f in the subinterval with the area of the rectangle of that height on the interval.

Define d(P)=max {xi-xi-1: i=1,...,n}. The Riemann integral exists iff there exists a limit for the values I(f;P;C) as d(P) -> 0. That is, there is some value I such that for any ε>0 there exists some δ>0 such that if d(P)<δ and we pick any points ci ∈ [xi-1,xi] we have that

|I-I(f;P;C)| < ε.
In such a case, I is the value of the Riemann integral.

In other words, we can guarantee that a Riemann sum will be a good approximation of the integral of a (Riemann-integrable!) function merely by forcing the partition to be fine enough.

In order to find the area underneath a curve, one can utilize a Riemann Sum. This involves slicing the region whose area needs to be found, and approximating the strips with rectangles.

The region is sliced into n strips, which partition the interval into n subintervals or increments. The x-values where the partition have taken place are termed the sample points. This will be the base of the rectangle. At each sample point, the corresponding function value is taken, and is treated as the altitude of the rectangle. Therefore, the area of any one rectangle is given by:

`Arect=f(xk)(Δxk), where k is 1, 2, 3, 4, all the way up to n`

Now, the integral, or the area underneath the curve, is approximately equal to the summation of all the rectangles. That is:

```         n
Area =   Σ   f(xk)Δxk
k=1
```

Now, there are several different types of Riemann Sums. One is termed the Lower Riemann Sum (L) which uses inscribed rectangles. Here, each sample point is taken so that f(x) is the lowest value in its respective subunit. One is the Upper Riemann Sum (U) which uses circumscribed rectangles. Here, each sample point is taken where f(x) is the highest. Usually, the area of the region underneath the curve will be somewhere between the Upper and Lower Sums. Here, is an illustration:

```
| \           /
9 |  \         /
8 |  |\       /
7 |  | \     /
6 |  |  \   /
5 |  |   ^-^
4 |  | R  |   Consider this function to be
3 |  |    |    f(x)=(x-3)2 + 5
--|--|----|-----------
0 1 2  3  4

Let us look for the area between x=1 and x=3.  The limits here are 1 and 3.  The minimum value of the function is at (3,5).
Now, if we want to find the Lower Sum then we should inscribe rectangles within the graph of the function.  First of all, we must decide on how many rectangles we want to use.  Let's use 4 rectangles, so we can get a good approximation.  Each rectangle will, therefore, have a base of 0.5 units.

|   \                      /       x       Δ(x)     f(x)      Δ(x)[f(x)]
9 |     \                  /        --------------------------------------
8 |     __\              /          1.5    |  0.5  |  7.25  |   3.625
7 |    |  |\           /              2    |  0.5  |     6  |       3
6 |    |  |__\       /              2.5    |  0.5  |  5.25  |   2.625
5 |    |  |  |_^ _ ^                  3    |  0.5  |     5  |     2.5
4 |    |  |  |  | |                --------------------------------------
3 |    |  |  |  | |                  By summing all the values in column 4,
--|----|--|--|--|-|--------          we find that the lower sum approximation
0   1     2    3    4             is 11.75.

This is the lower sum.

|   \ __                   /       x       Δ(x)     f(x)      Δ(x)[f(x)]
9 |    |\ |                /        --------------------------------------
8 |    |  \__            /            1    |  0.5  |     9  |    4.5
7 |    |  |\ |         /            1.5    |  0.5  |  7.25  |  3.625
6 |    |  |  \__     /                2    |  0.5  |     6  |      3
5 |    |  |  | ^|_ ^                2.5    |  0.5  |  5.25  |  2.625
4 |    |  |  |  | |                --------------------------------------
3 |    |  |  |  | |                  By summing all the values in column 4,
--|----|--|--|--|-|--------          we find that the upper sum approximation
0   1     2    3    4             is 13.75.

This is the upper sum.
```

Now we know that the area of the region under the curve is somewhere between 11.75 and 13.75. This is a nice approximation, but if we want an even closer approximation we would have to increase the number of rectangles. If we increase the number of rectangles, the altitudes of the rectangles in the lower sum model will increase, while the altitudes of those in the upper sum will decrease. Hence, we will be left with a smaller range. If we continue to increase the number of rectangles, or increments, the difference between the two will become very small. Using this idea, once the number of the rectangles reaches an infinite amount, the area will be exact. This is the precise idea behind integration. In other words:

```                   n
Area =   lim       Σ   f(xk)Δxk
Δx->0      k=1

or

n
Area =  lim         Σ   f(xk)Δxk
n->∞       k=1

or

b
Area =  ∫f(x)dx
a
```
This is the definition of integration.

Now, looking back to our example, if we want to find the exact area of the region, we can integrate the function using the limits we assigned previously of 3 and 1. Or, in other words:

```        3
Area =  ∫[(x-3)2 + 5]dx
1
```
The exact area is 12.6666 (repeating 6), which was between our upper and lower sums.

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