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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.

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*)

Answer.

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*), whereis 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 isheight ⋅ 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*) 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*(*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*) asThis 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.

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