Riemann Sums

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Riemann Sums
Field: Calculus
Image Created By: Marhot

Riemann Sums

A Riemann sum is an approximation of the area under a curve using a number of rectangles.

Basic Description

The area under a curve is used to measure how one quantity may change with respect to another quantity. For example, when a velocity curve is evaluated in terms of a change in time, the result will be a change in position for an interval of time (that is how an object changing position relates to time). Therefore, over that change of time, the change in position is evaluated and the displacement can be calculated. This value is represented by the area beneath the position curve.

Although the area under a curve is often measured using integration, there are various approximations that may be used to estimate this value. The Riemann sum, for example, fits one or more rectangles beneath a curve, and takes the total area of those rectangles as the estimated area beneath the curve. The order of the Riemann sum is the number of rectangles beneath the curve. Therefore, a single rectangle is a first order approximation while twelve rectangles, is a twelfth order approximation.

If more than a single rectangle is used, it is often most desirable to have the rectangles be the same width so only their lengths vary. This makes for quicker calculations, especially when the order is large, however, as noted in the more mathematical description, this is not always the most accurate method to approximate the area under a curve. If the curve peaks or its slope (the rate at which it increases or decreases in height), is not constant, unequal widths may allow for greater accuracy.


In this case, the widths a and b are equal, but the lengths of the two rectangles are different.

The rectangle's length is matched to the height of the curve at one of three points:


If the curve is below the x-axis, any of the rectangles concerning this portion are subtracted from the total area.


In this example, the total area is the area of the last four rectangles subtracted from the first two rectangles.

Overestimation and underestimation are two important concepts to address when approximating the area under a curve. Overestimation is when the rectangles approximate the area to be bigger than the actual area and underestimation is when the rectangles approximate the area to be less than the area beneath the curve. Overestimation occurs with a left Riemann sum and a curve that has a negative slope or with a right Riemann sum and a curve that has a positive slope. Underestimation occurs with a left Riemann sum and a curve that has a positive slope or a right Riemann sum and a curve that has a negative slope.


To avoid too much under or over estimation, it is sometimes best to use the the middle Riemann sum.

Note that Riemann sums are only an approximation or estimation, and the more rectangles that are used, the more accurate the estimation. This is because more of the points in the curve are accounted for. As one can see in the main image, the rectangles begin to shape more and more like the curve as more of them are added. Therefore, an infinite number of rectangles, meaning the width of each rectangle is infinitesimally small, is ideal; and this is what represents an integral.

A More Mathematical Explanation

A Riemann sum is a way to approximate the value of an an integral. This approximation is performed by [...]

A Riemann sum is a way to approximate the value of an an integral. This approximation is performed by fitting rectangles beneath the curve and adding their respective areas(length X width) together.

Typically, these summations will have rectangles of equal width with varying heights where the height of the rectangle meets the curve of concern at either the left endpoint of its base, the right endpoint of its base, or the midpoint of its base.

Therefore, if there are five rectangles, each with the same width,  w , and lengths, l_{1}, l_{2}, l_{3}, l_{4}, and l_{5}. Then the sum of the areas of these five rectangles is:

(w\cdot l_{1})+(w\cdot l_{2})+(w\cdot l_{3})+(w\cdot l_{4})+(w\cdot l_{5})

Which can be simplified to:


There is a general form of this summation in which each of the lengths are added together and multiplied by their equivalent width. The equation is:

S=\sum_{i=1}^{n}f(x_{i}^{*})(\Delta x)

Where f(x_{i}^{*}) is equal to the value of the function at the given value i, or the length of the rectangle, and n is the number of desired rectangles.

The calculation of this sum varies depending on whether it is a left, right, or middle sum. For the next three examples, assume that the interval in which the area is to be approximated starts at point  a .

For a left sum:

S=\Delta x(f(a)+f(a+\Delta x)+f(a+2\Delta x)+...+f(a+(n-1)\Delta x))

The equation starts at f(a) because this is the very first of the left endpoints of all the bases of the rectangles and ends at f(a+(n-1)\Delta x) since this is the last of the left endpoints. In this sum and the following two, multiples of \Delta x are added to  a to increment in the appropriate intervals. This is only true if all the widths of the rectangles are the same. If the widths were not all the same, then one would have to individually add each previous width to get to the next necessary end point.

For a right sum:

S=\Delta x(f(a+\Delta x)+f(a+2\Delta x)+f(a+3\Delta x)+...+f(a+n\Delta x))

This equation starts at  f(a+ \Delta x) because this is the first right endpoint while  f (a +n \Delta x) is the last.

For a middle sum:

S=\Delta x(f(a+\frac{\Delta x}{2})+f(a+2\frac{\Delta x}{2})+f(a+3\frac{\Delta x}{2})+...+f(a+n\frac{\Delta x}{2}))

This equation starts in between the left and right endpoints of the first base, which is why  \Delta x is divided by two. And this continues throughout since each midpoint is evaluated for the height of the curve.

Although having a constant \Delta x is easier for calculations, it is often more accurate to have unequal intervals, This increases accuracy for many functions that do not have constant slopes. For example:


As the slope increases, a smaller width in the rectangle allows for less overestimation in this specific approximation.

Where the slope increases, a smaller interval is better for approximating to avoid large amounts of over or under estimation.

Overestimation occurs with left Riemann sums with negative slopes because the rectangle’s height is measured to be higher than most of the part of the curve it is representing. While for a right Riemann sum, this would cause underestimation since it is mostly shorter than the rest of the represented portion of the curve. For curves with increasing slopes, the left Riemann sum would therefore compose rectangles of shorter height and the right Riemann sum would compose rectangles of greater height.

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http://www.math.hmc.edu/calculus/tutorials/riemann_sums/ http://awapcalculus11.wikispaces.com/file/view/4.3+Riemann+Sums+and+Definite+Integrals+(11).pdf http://www-rohan.sdsu.edu/~jmahaffy/courses/f00/math122/lectures/riemann_sums/riemanns.html http://en.wikipedia.org/wiki/Riemann_sum

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