Study Guide

# Definite Integrals - Comparing Right- and Left-Hand Sums

## Comparing Right- and Left-Hand Sums

Left Hand Sums and Right Hand Sums give us different approximations of the area under of a curve. If one sum gives us an overestimate and the other an underestimate,then we can hone in on what the actual area under the curve might be.

### Sample Problem

Let f be an increasing function on [a, b] and let R be the region between the graph of f and the x-axis on [a, b].

• Will LHS(n) be an over- or under-estimate of the area of R?
• Will RHS(n) be an over- or under-estimate of the area of R?

(hint: sketch f)

Whatever shape f has, we know f is increasing. This means on any sub-interval f will be smallest at the left endpoint and largest at the right endpoint of that sub-interval: Any left-hand sum will be an under-estimate of the area of R. Since f is increasing, a left-hand sum will use the smallest value of f on each sub-interval. The means any left-hand sum will fail to cover all of R. Any right-hand sum will be an over-estimate of the area of R. Since f is increasing, a right-hand sum will use the largest value of f on each sub-interval. This means any right-hand sum will cover R and then some. We see that if f is always increasing then a left-hand sum will give an under-estimate and right-hand sum will give an overestimate. If f is always decreasing then a left-hand sum will give an over-estimate and a right-hand sum will give an under-estimate.

If f alternates between increasing and decreasing, it's possible for both the LHS and RHS to be overestimates, or for both the LHS and RHS to be underestimates.

• ### Error in Left- and Right-Hand Sums

How far off are the left/right hand sums? It's sort of like thinking about how much a four-year-old colors outside of the lines with his spanking new, easy grip Crayola's.

If f is monotonic (either strictly increasing or strictly decreasing) on [a, b], then the area between f and the x-axis on [a, b] will be between LHS(n) and RHS(n). If f is strictly increasing, LHS(n) will be smaller and RHS(n) will be bigger. If f is strictly decreasing, RHS(n) will be smaller and LHS(n) will be bigger.

Either way, the area between f and the x-axis on [a, b] is between the area covered by the LHS rectangles and the area covered by the RHS rectangles.

We can use this to say roughly how far from reality our estimates with LHS(n) and RHS(n) are. Assume f is monotonic and let R be the area between f and the x-axis on [a, b]. Since we know the area of R is between LHS(n) and RHS(n), the area of R can't be farther from either LHS(n) or RHS(n) than those two are from each other.

### Sample Problem

Suppose f is increasing on [a, b] and we know that LHS(n) and RHS(n) are within 2 of each other. Then LHS(n) must be within 2 of the actual area, since

LHS(n) < Actual Area < RHS(n)

and we know RHS(n) is within 2 of LHS(n).

As n gets bigger, LHS(n) and RHS(n) get closer together. If we take n big enough, we can get LHS(n) and RHS(n) as close as we want. We can get LHS(n) and RHS(n) as close as we want to the real area of R, so long as we use enough rectangles.

To know how many rectangles we need, first we need a quick way to calculate the difference between LHS(n) and RHS(n). Finding each sum and then subtracting is too much work.

To find the difference between LHS(n) and RHS(n) on [a, b], take the difference between f(a) and f(b), then multiply by the width of a sub-interval.

Here's why the trick works. First assume f is increasing and nonnegative on [a, b]. Take RHS(n) and LHS(n), where is the width of a sub-interval.

We can look at the difference between RHS(n) and LHS(n) as the area that's covered by the little boxes "in between" the LHS and RHS rectangles. Now imagine scooting those little boxes over to make one big stack. The area of this stack is still the difference between RHS(n) and LHS(n).

The bottom of the stack is at f(a) and the top of the stack is at f(b). This means the height of the stack is (f(b) – f(a)). The width of the stack is Δx, since that's the width of a sub-interval.This means the area covered by the stack of boxes is

height ⋅ width = (f(b) – f(a)) ⋅ Δx.

This is the difference between the right-hand and left-hand sums.

If f is increasing then the difference between LHS(n) and RHS(n) is

(f (b) – f (a)) ⋅ Δx

and if f is decreasing then the difference between LHS(n) and RHS(n) is

(f (a) – f (b)) ⋅ Δx.

We can combine these two formulas into one. If f is monotonic, then the difference between LHS(n) and RHS(n) is

|f(b) – f(a)| · Δx.

This number must always be positive. Right now, we don't care if LHS(n) or RHS(n) is bigger. We care about how far apart they are.

We said that if we make n big enough (that is, use enough rectangles), we can get LHS(nand RHS(nas close as we want.

The difference between LHS(n) and RHS(n) is

|f(b) – f(a)|Δx.

Since we can write the difference between LHS(n) and RHS(n) as This quantity gets smaller as n gets bigger, since the values a, bf (a), and f (b) don't change when we take more rectangles. If we make n big enough, then we can get this quantity to be as small as we want.

## This is a premium product 