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# Definite Integrals

# 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 Δ x, 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)) ⋅ Δ*x*.

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)) ⋅ Δ*x*

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

(*f* (a)-*f* (b)) ⋅ Δ*x*.

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)|Δ*x*.

This number must always be positive. Right now, we don't care if *LHS*(*n*) and *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*(*n*) and *RHS*(*n*) as close as we want.

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

|*f* (b)-*f* (a)|Δ*x*.

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, b, *f* (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.