1. ## area under curve

Consider the limit of summed terms (attachment). Could someone please explain why each of the sums in the 'attached' expression gives an over-estimate of the area beneath y=x^2 +3, between x=0 and x=1.

****I'm rephrasing my question in LATEX so that more posters are attracted to assist me (please).

$\lim_{n \to 0} \frac {1}{n} \Sigma^{\infinity}_{k=1}\(\frac{k^2}{n^2} + 3)$

Could someone please explain why each of the sums in the above expression gives an over-estimate of the area beneath the curve $y = x^2 +3$ , between x=0 and x=1 *****

2. This is a Riemann Integral calculating the height of rectangle with the value of the function for their right side. That's why it overestimates. If you would have taken the left side you would have k going from 0 to n-1 instead of 1 to n and in that case it would underestimates.
Although, when you calculate the limit it goes to the same value. You thus only overestimates when there is no limit.

3. I'm rephrasing my question in LATEX so that more posters are attracted to assist me (please).

$\lim_{n \to 0} \frac {1}{n} \Sigma^{\infinity}_{k=1}\(\frac{k^2}{n^2} + 3)$

Could someone please explain why each of the sums in the above expression gives an over-estimate of the area beneath the curve $y = x^2 +3$ , between x=0 and x=1

4. Originally Posted by vincisonfire
This is a Riemann Integral calculating the height of rectangle with the value of the function for their right side. That's why it overestimates. If you would have taken the left side you would have k going from 0 to n-1 instead of 1 to n and in that case it would underestimates.
Although, when you calculate the limit it goes to the same value. You thus only overestimates when there is no limit.
Sorry vincisonfire, but I haven't come across Riemann Integrals before, could you briefly explain it?

Thanks

5. Originally Posted by vincisonfire
This is a Riemann Integral calculating the height of rectangle with the value of the function for their right side. That's why it overestimates. If you would have taken the left side you would have k going from 0 to n-1 instead of 1 to n and in that case it would underestimates.
Although, when you calculate the limit it goes to the same value. You thus only overestimates when there is no limit.
Hey vincisonfire, could you please explain it in simpler terms?

thanks

6. Can anyone else voice their opinion on this question?

7. tsal15,

It's not really easy to explain without drawings, so please refer to this link if you feel you need graphical visualizations.

A Riemann Integral is a definition of the area 'under a curve'. Imagine that you lived in a world with no calculus, and you really really needed to know at least an approximation of the area under a curve.

Assuming you do know how to calculate the area of rectangles, you'd eventually think about discretizing the curve. That is, splitting it into lots of rectangles, calculate the area of each and then sum all of them.

Let $f(x)$ be your curve and $a,b$ be your area bounds. If you have $n$ rectangles, each rectangle will have a base of $\frac{b-a}{n}$. Now you need to define arbitrarially what is the height of your rectangle going to be. Now it's up to you: you can take the leftmost value of $f(x)$ in your rectangle interval, or the rightmost, or the middle, or any other arbitrary value relative to your rectangle. Notice that if $f(x)$ is crescent, taking the rightmost value will overestimate your rectangle, and thus your total area. I hope you can visualize it: the rectangle will 'overflow' the function in the top.

Obviously, as $n$ grows larger, your area tends to the real area under the curve. That's why the value of the limit when $n\to\infty$ is defined to be the real area under the curve, according to the Riemann definition.

Notice that the Riemann Integral is INITIALLY unrelated to antiderivatives, or the indefinite integrals. In particular, the indefinite integral is a function, and the Riemann Integral is a number. They are connected by the Fundamental Theorem Of Calculus, as you may or may not know (yet).

Hope this helps,

8. Thanks rafael almeida, i think i get it now.

also how do i evaluate $lim_{n \to infinity} \frac {1}{n} \Sigma_{k=1} ^{n} \(\frac {k^2}{n^2} +3)$
using the expression:

$\Sigma_{k=1} ^{n} \k^2 = \frac {1}{6} n(n+1)(2n+1)$

I tried it myself and evaluated it to zero... i'm pretty sure this is wrong...could u show me the correct way thanks

9. Originally Posted by tsal15
Thanks rafael almeida, i think i get it now.

also how do i evaluate $lim_{n \to infinity} \frac {1}{n} \Sigma_{k=1} ^{n} \(\frac {k^2}{n^2} +3)$
using the expression:

$\Sigma_{k=1} ^{n} \k^2 = \frac {1}{6} n(n+1)(2n+1)$

I tried it myself and evaluated it to zero... i'm pretty sure this is wrong...could u show me the correct way thanks
You want to evaluate $L = \lim_{n\to\infty} \frac {1}{n} \sum_{k=1} ^{n} \left(\frac{k^2}{n^2} + 3\right)$ using the fact that $\sum_{k=1} ^{n} k^2 = \frac{1}{6} n(n+1)(2n+1)$. (hint: click on the equations to see the LaTeX I used to form them)

My approach would be:

$\lim_{n\to\infty} \frac {1}{n} \sum_{k=1} ^{n} \left(\frac{k^2}{n^2} + 3\right) = \lim_{n\to\infty} \frac {1}{n} \sum_{k=1} ^{n} \left(\frac{k^2}{n^2}\right) + \lim_{n\to\infty} \frac {3n}{n} = \lim_{n\to\infty} \frac {1}{n^3} \sum_{k=1} ^{n} (k^2) + 3$

And then apply your formula for the sum of the first n squares:

$\lim_{n\to\infty} \left(\frac{n(n+1)(2n+1)}{6n^3}\right) = \lim_{n\to\infty} \left(\frac{2n^3 + 3n^2 + n}{6n^3}\right) = \lim_{n\to\infty} \left(\frac{n^3\left(2 + \frac{3}{n} + \frac{1}{n^2}\right)}{6n^3}\right)$

And you can safely cross out the $n^3$ term because as $n \to \infty$, it is far away from zero. Also, the terms $\frac{3}{n}$ and $\frac{1}{n^2}$ go to zero when $n \to \infty$. So this leaves you with:

$\lim_{n\to\infty} \left(\frac{n^3\left(2 + \frac{3}{n} + \frac{1}{n^2}\right)}{6n^3}\right) = \lim_{n\to\infty} \left(\frac{2}{6}\right)$

So, finally,

$L = \frac{2}{6} + 3 = \frac{10}{3}$

I hope I have not made major mistakes. However, this is the method.

Hope this helps,