How Do You Calculate Riemann Sums for sqrt(x) with a Squared Partition?

[fishingspree2]

Hello everyone, I have passed my integral calculus class and it's been a little while so I don't really remember everything. Can anyone help me out with this?

Homework Statement

Let f(x) = sqrt(x), x E [0,1]

and $$P=\left \{ 0,\left ( \frac{1}{n} \right )^{2}, \left ( \frac{2}{n} \right )^{2}...\left ( \frac{k-1}{n} \right )^{2}, \left ( \frac{k}{n} \right )^{2}...\left ( \frac{n-1}{n} \right )^{2}, 1\right \}$$ a partition of [0,1]

a) Find $$\Delta x_{k}$$ and $$f\left ( x_{k} \right )$$

The Attempt at a Solution

I don't understand the question. I remember the general method, we divide the interval in n little parts and we let n go to infinity, the parts get smaller and smaller and we add their area to find the total area over the interval. but i don't understand what's asked. Why are the elements in the partition squared?

thank you

 

[grief]

This is a specific partition of [0,1] (notice that the points are not evenly spaced, but they still form a partition). I think all you're asked to do is find the distances (the deltas) between the points of the partition, as well as the function values of the points of the partition. This question doesn't seem to be asking about any limits.

 

[fishingspree2]

how do i do that can you please give me a hint :shy:

 

[sutupidmath]

Remember, how do you partition an arbitrary interval [a,b]? What is $$\triangle x_k=x_k-x_{k-1}$$ right?
What does x_k represent, the k-th partition point right? which one is that in P?

 

[fishingspree2]

the length of one interval is $$\left ( \frac{k}{n} \right )^{2}-\left ( \frac{k-1}{n} \right )^{2}=\frac{2k-1}{n^{2}}$$

is the area under the curve on the interval [0,1] $$\lim_{n\to \infty}\sum_{i=1}^{n}\left ( \frac{2k-1}{n^{2}} \right )\cdot f\left ( \frac{2k-1}{n^{2}} \right )$$? i don't know what to do with the k in that expression

 

[sutupidmath]

Remember it is not $$f(\triangle x_k)..but...f(x_k)$$

 

[fishingspree2]

Alright, this is what i have found (remember that f(x) = sqrt(x)

$$\sum_{k=1}^{n}f(x_{k})\Delta x_{k}=\frac{1}{2}n+1-\frac{1}{2n^{2}}$$

now if I take the limit as n goes to infinity, I get infinity... I don't understand