Use Upper And Lower Sums To Evaluate An Integral

[pollytree]

Homework Statement

The question is to use upper and lowers sums, U_n and L_n, on a regular petition of the intervals to find the integral from 1 to 2 of f(x) = [[x]], where [[x]] is the greatest integer function.

Homework Equations

$$\Delta$$x = $$\frac{b-a}{n}$$

The Attempt at a Solution

$$\Delta$$x = $$\frac{2-1}{n}$$ = $$\frac{1}{n}$$

The minimum and maximum of f(x) on every subinterval of [1,2] would be 1 except for the subinterval which includes x=2 where the maximum value of f(x)=2, so for f(x) there exists a unique number I such that L_n $$\leq$$ I $$\leq$$ U_n

I know that the area will be 1, as for [1,2) U_n = L_n = $$\Delta$$x$$\sum$$1 = $$\frac{1}{n}$$ x 1n = 1

However I am not sure how to include the subinterval that contains x=2 into my calculations. Any help on this would be great!

Thanks!

 

[Office_Shredder]

If you have n subintervals, evenly spaced say, then n-1 will have f(x) be just 1, and one will contain a point where f(x) is 2. So your lower sum, using the lowest value on each interval, will just give you the area of 1 as expected. For the upper sum, you have

$$(\sum_{i=1}^{n-1} \frac{1}{n})+\frac{2}{n}$$ adding up all the parts where the upper bound is 1, plus the one part where the upper bound is 2. Can you find the limit of that summation as n goes to infinity? (Of course we know it should be 1)

 

[pollytree]

That makes perfect sense! Thanks a bunch :D