Find lower and upper estimates of integral

Hi, my question is:

Consider the integral of 1/(t(t+1)) with respect to t, from lower limit t=0 to the upper limit t=n, where n is an arbitrary positive integer.

Let P be the partition {1, 2, ..., n}. Find lower and upper estimates L(P) and U(P) for the integral above. Simplify your answer as far as possible.

I know that you need to use Riemann's integral but I'm really stuck on this question, so any help would be greatly appreciated! :)

Re: Find lower and upper estimates of integral

Take the greatest and the smallest value of 1/(t(t+1)) on each interval [k, k + 1] for k = 0, ..., n - 1. Multiply these values by the width of the segment, i.e., 1, then add the results to get the upper and lower estimates.

Re: Find lower and upper estimates of integral

This is my rough working:

L(P)= sum from i=1 to n of g(t(i-1))(t(i)-t(i-1))

t(i)-t(i-1)=(n-1)/n

g(t(i-1))=1/(t(i-1))(t(i-1)+1)

Then if I put these back into the equation for L(P), I should get the lower estimate. However, I end up getting a complicated summation, and I'm not sure about how to simplify it, or if it's even correct. Here is the summation that I get when finding L(P):

sum from 1 to n of (n(n-1))/(t(i-1)(t(i-1)+1))

Where t(i-1)=1+i(n-1)/n

Could you tell me if I'm on the right lines here, or if I'm completely incorrect? And if it's completely wrong, could you please help me? Thank you!

Re: Find lower and upper estimates of integral

Quote:

Originally Posted by

**sakuraxkisu** This is my rough working:

L(P)= sum from i=1 to n of g(t(i-1))(t(i)-t(i-1))

t(i)-t(i-1)=(n-1)/n

g(t(i-1))=1/(t(i-1))(t(i-1)+1)

Could you tell me if I'm on the right lines here, or if I'm completely incorrect? And if it's completely wrong, could you please help me?

Look at the graph. It is a decreasing function.

Therefore, an upper sum is a left-hand sum and lower sum is a right-hand sum.

Re: Find lower and upper estimates of integral

Quote:

Originally Posted by

**sakuraxkisu** Where t(i-1)=1+i(n-1)/n

Another thing is that this t(i) does not generate the partition {1, 2, ..., n} because (n-1)/n is not an integer for n > 1.

Re: Find lower and upper estimates of integral

Thank you for the tips, but to be honest I'm still a bit confused ..... Sorry to be a pain! Plato, I'm not really sure about what you mean by left hand sum and right hand sum, and Emakarov, would t(i-1) just be 1 in this case? Thanks again :)

Re: Find lower and upper estimates of integral

Quote:

Originally Posted by

**sakuraxkisu** This is my rough working:

L(P)= sum from i=1 to n of g(t(i-1))(t(i)-t(i-1))

Here for each i, you consider the segment [t(i-1), t(i)]. You multiply the value of g at the *left* end of this segment (i.e., g(t(i-1))) by the length of the segment and add the results. However, as Plato said, the function is decreasing, so the value at the left end of the segment is the largest possible value at that segment. So you get the upper and not the lower estimate. That's what Plato meant by saying "a lower sum is a right-hand sum".

Quote:

Originally Posted by

**sakuraxkisu** would t(i-1) just be 1 in this case?

If you want to find a formula for t(i), it should be such as to produce values in the partition, i.e., 1, 2, ..., n for various values of i. So t(i) can certainly not equal 1 for all i.

Quote:

Originally Posted by

**sakuraxkisu** t(i)-t(i-1)=(n-1)/n

Here you probably assume that since the partition is {1, 2, ..., n}, it has n intervals, and their total length is n - 1. But there are n points, so the number of intervals is n - 1, and the length of each interval is (n-1) / (n-1) = 1, which is obvious without calculating the total length and number. Conforming to the definition of the Riemann sum in Wikipedia, we could define t_{i} = i + 1, so that t_{0} = 1 and t_{n-1} = n (the largest index equals the number of intervals). Then the lower (and right) Riemann sum is

$\displaystyle g(2)(2-1)+g(3)(3-2)+\dots+g(n)(n-(n-1)) = \sum_{i=1}^{n-1}g(t_i)(t_i-t_{i-1}) =$ $\displaystyle \sum_{i=1}^{n-1}g(i+1)\overset{k=i+1}{=} \sum_{k=2}^ng(k)= \sum_{k=2}^n\frac{1}{k(k+1)}$.

Re: Find lower and upper estimates of integral

It should noted that $\displaystyle \int_0^n {\frac{{dt}}{{t(t + 1)}}} $ **diverges**.

The function $\displaystyle g(t)=\frac{{1}}{{t(t + 1)}}} $ is not bounded on $\displaystyle [0,1]$

I suspect that it is meant to be $\displaystyle \int_1^n {\frac{{dt}}{{t(t + 1)}}} $.

Then the set $\displaystyle \{1,2,\cdots n\}$ makes sense as a partition for $\displaystyle [1,n]$.

Then $\displaystyle g(t) = \left[ {\frac{1}{t} - \frac{1}{{t + 1}}} \right]$.

A left-hand sum is $\displaystyle \sum\limits_{k = 1}^{n - 1} {g(k)(1)} =~ ?$

A right-hand sum is $\displaystyle \sum\limits_{k = 2}^{n } {g(k)(1)} =~ ?$

Re: Find lower and upper estimates of integral

Thank you for your help! Sorry about that Plato, I meant t=1 as the lower limit but I entered the wrong number! Also, Emakarov, I can't seem to find any way to simplify the summation further, I was wondering, can it be simplified further? Thanks again for your help :)

Re: Find lower and upper estimates of integral

Quote:

Originally Posted by

**sakuraxkisu** I can't seem to find any way to simplify the summation further, I was wondering, can it be simplified further?

Yes, it can. Use the representation of g(t) as a difference from post #8. L(P) is a telescoping sum.

Re: Find lower and upper estimates of integral

Thank you so much for your help both of you! Sorry to have been such a pain!