A Riemann sum is a method of evaluating definite integrals; it is in my opinion an intuitive idea, using rectangular strips from the x axis to approximate the area under the curve. At a finite level this only provides an approximate of the area under the curve, but as the strip length of the rectangles tends to $0 \ (\delta x \to 0)$ it ceases to be an approximate and provides an exact area.
Naturally an example is the best way to visualise this, consider the function: \[ f(x) = x^2 \ \text{between} \ 1 \le x \le 4 \] As this closed interval is of size 3 this is an obvious number of strips to use. I'm taking the height of each rectangle to be the value of $f(x)$ at the right most point of the rectangle. This may seem confusing at the moment but the graphics below should make it all clear.
Ta-da! Now as there are a finite number of rectangles (3) this is only an approximate of the area under the curve. Suppose I wanted a better approximation, I could try 9 rectangles.
I could better my approximation again by using 27 rectangles.
I could keep on improving my approximates by increasing the number of rectangles further. However it should now be clear that increasing the number of rectangles as well as decreasing the width of the rectangles improves the approximation. And as I said earlier as the width tends to $0$ and the the number of rectangles using tends to infinity this becomes the exact area under the curve.
Riemann sums come in three flavours: the left sum, the right sum and the middle sum. The prefix refers to the point of the rectangle that is taken up to the curve $f(x)$. So in my example it was the right, however you could use the left most point or the mid point of the rectangle instead. The most accurate of these for use in approximations is the middle sum.
Formulating a Riemann sum is not particularly difficult either. Consider the size of an individual rectangle, it's just $f(x) \times \Delta x$ where $\Delta x$ denotes the width of an individual rectangle and $f(x)$ the height at the point $x$. When the interval of integration between $a$ and $b$ is split into uniformly wide pieces it becomes clear that: \[ \Delta x = \frac{b-a}{n} \] Each rectangle is of a different height based on its position on the curve $f(x)$ and so each successive height must be taken an additional $\Delta x$ further ahead, or more mathematically as: \[ x_k = a + k \Delta x \] From this we can formulate the right Riemann sum: \[ S = \ \sum_{k=1}^{n} f(x_k) \ \Delta x \] For the right sum the height of each rectangle $f(x)$ is taken at the leading edge of the rectangle.
The other two Riemann sums can be formulated in a similar way, however for finding exact areas they are less useful so I have not included them.
To do this just put numbers into the formulae. I'll use the first approximate using 3 strips and the right summation as an example. So: \[ \Delta x = \frac{4-1}{3} = 1\] The formulated sum is: \[ S = \sum_{k=1}^{3} (1 + k \times 1 )^2 \times 1 \] Or in its more simplified form: \[ S = \sum_{k=1}^{3} ( 1 + k )^2 \] Evaluating this yields: \[ S = 29\] In this case the approximate is fairly close to the actual value of 21, however for larger more complex functions such a small number of strips usually provides a poor estimate.
For the purposes of finding definite integrals the right sum is the best choice as summations from $1 \ \text{to} \ n$ have forms to work with and make it possible to find exact solutions using Riemann sums.
As n tends to infinty the strip length $\Delta x \to 0 $ When we have a summation of an infinite number of strips in a given interval we call this an integral. At its most basic level an integral is just a summation of all the strips in a range. \[ \lim_{n \to \infty} \ \sum_{k=1}^{n} f(x_k) \ \Delta x \equiv \int_a^b f(x) \ \mathrm{d}x \] Now you might ask, why is this the case, well this is how we define the riemann integral. And it makes sense intuitively too, the definite integral is just the area under the curve, which is exactly what this summation represents. And for this summation the Riemann sum looks quite different, it is an exact representation of the area under the curve.
These can be shown to be equivilent, consider the function: \[ f(x) = x^3 \text{between} \ 1 \le x \le 4 \]
So for the standard method of integration: \[S = \int_{1}^{4} x^3 \mathrm{d}x\] Evaluating this: \[\begin{align} S &= \Big[ \frac{x^4}{4} + C \Big]_{1}^{4} \\ S &= 64 - \frac{1}{4} \\ S &= \frac{255}{4} \end{align} \]
The Riemann sum is a bit harder to do, it is also quite tedious to perform as even fairly simply integrals will throw up fairly nasty summations that are difficult to simplify down by hand.
So in this summation the strip length will be: \[ \Delta x = \frac{4-1}{n} = \frac{3}{n} \] and the position on the $x$ axis is given by: \[ x_k = 1 + \frac{3k}{n} \] So we will be evaluating the Riemann sum: \[ S = \lim_{n \to \infty} \ \sum_{k=1}^{n} (1+x)^3 \ \Delta x \] $\Delta x$ can be factored out because it's a constant: \[ S = \lim_{n \to \infty} \ \frac{3}{n} \sum_{k=1}^{n} \Big( 1+\frac{3k}{n} \Big)^3 \] Expanding the brackets yields: \[ S = \lim_{n \to \infty} \ \frac{3}{n} \sum_{k=1}^{n} \Big( \frac{27k^3}{n^3} + \frac{27k^2}{n^2} + \frac{9k}{n} +1 \Big) \] This can be broken down into four individual summations ($\Delta x$ is included in the term that is factored out): \[ S = \lim_{n \to \infty} \ \frac{81}{n^4}\sum_{k=1}^{n} k^3 + \frac{81}{n^3} \sum_{k=1}^{n} k^2 + \frac{27}{n^2} \sum_{k=1}^{n} k + \frac{3}{n} \sum_{k=1}^{n} 1 \] Evaluating each summation yields: \[ S = \lim_{n \to \infty} \ \frac{81}{4n^2}(n+1)^2 + \frac{81}{6n^2}(n+1)(2n+1) + \frac{27}{2n}(n+1) + 3 \] Simplifying this yields: \[ \begin{align} S &= \lim_{n \to \infty} \ \frac{255}{4} + \frac{189}{2n} + \frac{135}{4n^2} \\ \\ S &= \frac{255}{4} \end{align} \] Which is the same as the more standard method.
Riemann sums formalise the method of finding the definite integral. As you saw above they are usually fairly complex and require a lot of work, similar to finding derivatives from first principles. The Riemann sum really has little use outside of defining the definite integral, shortcut methods of integration are much easier to remember and perform. However Riemann sums provide a point to definite integration from in much the same way infinitesimals do for differentiation.
*All the graphics used in this were made by myself using Mathematica 7.0, if you wish to use them please provide credit and link back to my blog.
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