# Right-Hand Sum

There is much debate about who is more awesome...right-handers or south-paws. Would you want Shoeless Joe Jackson on your team, or Nomar Garciaparra? That's why it's better to have a switch hitter like Chipper Jones at bat; we don't have to pick.

In calculus, right-hand sums are similar to left-hand sums. However, instead of using the value of the function at the left endpoint of a sub-interval to determine rectangle height, we use the value of the function at the *right endpoint* of the sub-interval.

As with left-hand sums we can find the values of the function that we need using formulas, tables, or graphs.

## Right-Hand Sums with Formulas

If we're given a formula for the function, we can use this formula to calculate the value of the function at the right endpoint of each sub-interval.

## Right-Hand Sums with Tables

In order to find a right-hand sum we need to know the value of the function at the right endpoint of each sub-interval. We can take a right-hand sum if we have a table that contains the appropriate function values.

### Sample Problem

Some values of the decreasing function *f* ( *x* ) are given by the following table:

- Use a Right-Hand Sum with 2 sub-intervals to estimate the area between the graph of
*f*and the*x*-axis on the interval [0,4].

We don't know what the function*f*looks like, but we know these points are part of it:

Dividing the interval [0,4] into 2 equally-sized sub-intervals gives us sub-intervals of length 2. The height of the rectangle on [0,2] is *f** *(* 2 *) = 17, so the area of this rectangle is

height ⋅ width = 17(2) = 34.

The height of the rectangle on [2,4] is *f* (4) = 3, so the area of this rectangle is

height ⋅ width = 3(2) = 6.

Adding the areas of these rectangles, we estimate the area between the graph of *f* and the *x*-axis on [0,4] to be

34 + 6 = 40.

- Use a Right-Hand Sum with 4 sub-intervals to estimate the area between the graph of
*f*and the*x*-axis on the interval [0,4].

Answer. Dividing the interval [0,4] into 4 evenly-sized sub-intervals produces sub-intervals of length 1.

Sub-interval [0,1]: This rectangle has height *f* ( *0* ) = 20 and width 1, so its area is 20.

Sub-interval [1,2]: This rectangle has height *f* (1) = 18 and width 1, so its area is 18.

Sub-interval [2,3]: This rectangle has height *f** *( 2 ) = 17 and width 1, so its area is 17.

Sub-interval [3,4]:This rectangle has height *f* (3) = 11 and width 1, so its area is 11.

Adding the areas of these rectangles, we estimate the area between the graph of *f* and the *x*-axis on [0,4] to be

20 + 18 + 17 + 11 = 66.

- Are the estimates in parts (b) and (c) over- or under-estimates for the area between the function
*f*and the*x*-axis on the interval [0,4]?

Answer. We don't know what the function*f*looks like exactly, but we know it's a decreasing function that passes through these points.

That means it must look something like this:

Our estimates in (b) and (c) were both underestimates, because the rectangles didn't cover all of the area between the graph of *f* and the *x*-axis on [0,4]:

- Could we use a right-hand sum with more than 4 sub-intervals to estimate the area between the function
*f*and the*x*-axis on the interval [0,4]?

Answer. No. The table doesn't contain enough data for us to divide the interval [0,4] into more than 4 sub-intervals. If we tried to use 8 sub-intervals, for example, we would need to know *f* (0.5), and that value isn't in the table.

## Right-Hand Sums with Graphs

When finding a right-hand sum, we need to know the value of the function at the right endpoint of each sub-interval. We can find these values by looking at a graph of the function.

## Right-Hand Sum Calculator Shortcuts

For a LHS, we only use values of the function at left endpoints of subintervals. We never use the value of the function at the right-most endpoint of the original interval.

For Right-Hand sums, it's the other way around. For a RHS we only use values of the function at right endpoints, so we'll never use the value of the function at the left-most endpoint of the original interval.

## Right-Hand Sums with Math Notation

After learning the notation for left-hand sums, the notation for right-hand sums requires a very slight adjustment. Assume that we're using sub-intervals all of the same length and we want to estimate the area between the graph of *f* ( *x* ) and the *x*-axis on the interval [a,b].

An interval of the form [a,b] has length ( *b* – *a* ). If we wish to divide the interval [a,b] into *n* equal sub-intervals, each sub-interval will have length

The endpoints of the sub-intervals are

To take a RHS with *n* sub-intervals we find the value of *f* at every endpoint but the first, add these values, and multiply by Δ*x*.

*RHS*(*n*) = [*f* (*x*_{1}) + *f* (*x*_{2}) + ... + *f* (*x*_{ n – 1}) + *f* (*x*_{n})]Δ*x*

Summation notation for extra fanciness is optional:

## Right-Hand Sums with Sub-Intervals on Different Lengths

As with left-hand sums, we can take right-hand sums where the sub-intervals have different lengths.

## Sample Problem

Values of the function *f* are shown in the table below. Use a right-hand sum with the sub-intervals indicated by the data in the table to estimate the area between the graph of *f* and the *x*-axis on the interval [1,8].

Answer. The sub-intervals given in this table aren't all the same. Most of them are 2, but one is 1.

On sub-interval [1,3] the height of the rectangle is *f* (3) = 5 and the width is 2, so the area is

5(2) = 10.

On sub-interval [3,4] the height of the rectangle is *f* (4) = 3 and the width is 1, so the area is

3(1) = 3.

On sub-interval [4,6] the height of the rectangle is *f* (6) = 5 and the width is 2, so the area is

5(2) = 10.

On sub-interval [6,8] the height of the rectangle is *f* (8) = 1 and the width is 2, so the area is

1(2) = 2.

Adding the areas of the rectangles, we estimate the area between *f* and the *x*-axis on [1,8] to be

10 + 3 + 10 + 2 = 25.