Summation Notation and Definite Integrals

General Form of Summation Expressions To evaluate the integral using summation notation and letting the number of rectangles go to infinity (thus each rectangle will be infinitely thin), the following patterns are used: Left-endpoints Right-endpoints

The definite integral is directly linked to the summation of infinitely many, infinitely thin rectangles. Consider the case where we divide the interval [a, b] into n-subintervals of equal width and build rectangles up to the curve f(x) from the x-axis. For simplicity, assume the curve has no negative values on this interval. If we allow for n rectangles of equal width, how do we find the area of a single rectangle?

Height × Width: , where .

The trick is to determine which x-values to "grab" and to then find the associated f(x), to serve as the height of each rectangle. Let us call the selected x-values x_i and the height of the rectangles f(x_i). If we use the x-values at the left-endpoints of these intervals to calculate height, we need to start the first rectangle at a (the left bound of the entire interval). The second will be one width of a sub-interval over from a, the next will be two subintervals over from a, and so on, until the last is one subinterval to the left of b. Fortunately, there is an efficient way to count along in the pattern 1, 2, 3, … in mathematics: use a summation.

If we wanted to add the x-values in an interval (using left-endpoints of the subintervals), we would first need to ensure that we started at the left-endpoint of the entire interval (a) and then steadily march through the entire interval, getting an x-value for each subinterval. To accomplish this, we would use the following summation:

We do not want to add just the x-values, however, we need the f(x) associated with each, thus we need the following summation to sum together all the the "y-values":

To find the area of n rectangles, each f(x) value (representing height) needs to be multiplied by the associated width, which is (if the rectangles are all have the same width). Hence, the sum of the area of n rectangles is:

Finally, to allow for the summation of the rectangles to produce the exact area of the region in question, let the number of rectangles go to infinity (forcing each to be infinitely thin), yielding the following summation (using left-endpoints):

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If we decided to use right-endpoints for the height of each rectangle, the x-values would be chosen in the following pattern: The first would start at one width of a subinterval to the right of the left-endpoint of the entire interval, the next one subinterval beyond that, and so on, ending with the right-endpoint of the entire interval.

Thus, if we wanted to add the x-values in an interval (using right-endpoints of the subintervals), we would use the following summation:

How do we add all the y-values and then multiply the total by the width (which is the same for all rectangles) using right-endpoints? The answer to that question is the following summation:

To find the exact area, let the number of rectangles go to infinity (thus each rectangle will be infinitely thin).

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Examples:

Set up the summation that will calculate the following integral:

By left-endpoints on [0, 1]:

By right-endpoints on [0, 1]:

Solving this summation series using the known formula for "i-squared" proceeds as follows:

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Now changing the interval so that it does not start at zero, note the changes in the summation series that are necessary to evaluate , whose total interval width is 3, starting at -1 and ending at 2:

By left-endpoints on [-1, 2]:

By right-endpoints on [-1, 2]:

A table can be created to demonstrate that on [-1, 2] using 6-subintervals, the summation formulas select the correct x-values:

i Left-endpoints Right-endpoints 1 –1.0 –0.5 2 –0.5 0.0 3 0.0 0.5 4 0.5 1.0 5 1.0 1.5 6 1.5 2.0

A calculator handles this "check" more efficiently:

Left-endpoints

Right-endpoints

  

The following calculator screens show how the series, with n = 6, 20, and 200, seem to be converging on the value of the integral.

  

Does the series actually converge to exactly 3? Yes. This can be proved using the summation formula for "i-squared," as was shown above with the interval [0, 1].

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Additional examples:

Using left-endpoints:

Using right-endpoints:

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A Question of Proof

The key question to all these problems is whether or not we can prove the series will converge to the numerical value produced by the integration process. Proofs of the convergence (or divergence) of series are the central focus of the branch of mathematics called "real analysis." Just as the ability to find limits was the key to finding derivatives, finding limits of series is the key to proving why integration works.

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AP Style Questions:

This series could be tied to multiple choices. The series represents a Riemann Sum, using right-endpoints, from the integral with width 3, from 1 through 4. The evaluation of this integral yields:

Is the Riemann Sum, using n rectangles, evaluated at right-endpoints, on the interval from 1 through 4, going to be larger or smaller than the true value of the integral?

Here you would need to determine if the function is increasing or decreasing on the interval. Increasing or decreasing is a first derivative question (easily calculated). On [1, 4], this function is decreasing; hence right-endpoint Riemann Sums yield an approximation that is below the true value of the integral.

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