Right Endpoint Sums

Building a Right-Endpoint Sum

A Riemann sum estimates the accumulation represented by a collection of rectangles. To construct one, divide an interval into subintervals, choose a sample point in each subinterval, and multiply the function value at that sample point by the subinterval’s width. A right-endpoint sum uses the right edge of each subinterval as its sample point.

The distinction between left and right endpoints is simple, but it changes the rectangle heights and therefore the sum. The method is especially systematic when the interval is divided into equally wide subintervals.

The sum has \(n\) terms because there are \(n\) subintervals and one rectangle for each subinterval. The index begins at \(i=1\): \(x_1\) is the right endpoint of the first subinterval, and \(x_n=b\) is the right endpoint of the last one.

Why the Formula Uses These Function Values

On subinterval \([x_{i-1},x_i]\), a right-endpoint rectangle has width \(x_i-x_{i-1}=\Delta x\). Its height is the function value at the right edge, \(f(x_i)\). Thus its signed contribution is \(f(x_i)\Delta x\). Adding the contributions from all \(n\) rectangles gives the right-endpoint sum.

Summing these contributions for \(i=1,2,\ldots,n\) produces \(R_n=\sum_{i=1}^{n}f(x_i)\Delta x\). If the function is negative at a sample point, the corresponding rectangle contributes a negative amount. This is a signed sum, consistent with signed accumulation.

A reliable setup keeps three quantities distinct: the number of subintervals \(n\), the width \(\Delta x\), and the right endpoint \(x_i\) at which each function value is evaluated. In particular, the right endpoint is not the index \(i\) itself; it is \(a+i\Delta x\).

A Repeatable Setup

Worked Examples

Right Endpoints and Indexing

For equal subintervals, the \(i\)th left endpoint is \(x_{i-1}=a+(i-1)\Delta x\), while its right endpoint is \(x_i=a+i\Delta x\). This gives a useful comparison:

The width is unchanged; only the sample points change. For a right-endpoint sum, the list of sample points excludes \(a\) and includes \(b\). This is a quick way to check whether the indexing matches the requested method.

Common Mistake: Using the Left Endpoints

Consider \(f(x)=x^2\) on \([0,2]\) with \(n=4\). The width is \(\Delta x=\frac{2-0}{4}=\frac12\). A mistaken calculation might use \(0,\frac12,1,\frac32\), giving

This calculation uses the left endpoint of each subinterval, so it is a left-endpoint sum, not a right-endpoint sum. For the right-endpoint sum, the sample points must instead be \(\frac12,1,\frac32,2\):

The correction is not to change the width or the number of terms. It is to use the right edge of each subinterval when choosing every height.

When a Right-Endpoint Sum Is Useful

A right-endpoint sum turns sampled function values into a finite estimate of signed accumulation. If \(f\) describes a rate, each product \(f(x_i)\Delta x\) estimates the amount accumulated over one subinterval, and the sum combines those estimates. Units provide a useful check: a rate measured in meters per second multiplied by a time width in seconds gives meters.

The sum is an approximation based on the selected partition and sample points. It should not be confused with an exact accumulated value merely because the arithmetic is exact. For the calculations here, the stated sum is the exact result of the chosen rectangle method; that does not by itself establish that it equals the total accumulation represented by the function.

Before finalizing a calculation, check that:

• the interval endpoints and number of subintervals give the stated \(\Delta x\);
• there are exactly \(n\) sample points;
• the points begin one width to the right of \(a\) and end at \(b\);
• each function value is multiplied by the width; and
• the final units agree with height multiplied by width, when quantities have units.