Integral as a sum

1. Apr 26, 2017

Karol

1. The problem statement, all variables and given/known data

Why specifically 1/2 is the coefficient in CK? the sum, basically, doesn't change except for the coefficient. i can choose it as i want.
I understand the sum must equal the integral but i guess that's not the reason

2. Relevant equations
Area under a curve as a sum:
$$S_n=f(c_1)\Delta x+f(c_2)\Delta x+...+f(c_n)\Delta x$$

3. The attempt at a solution
$$f(c_1)\Delta x=\frac{x_k-x_{k-1}}{\frac{\sqrt{x_{k-1}}+\sqrt{x_k}}{2}}=...=2(\sqrt{x_k}-\sqrt{x_{k-1}})$$
$$S_n=2(\sqrt{x_1}-\sqrt{x_0}+\sqrt{x_2}-\sqrt{x_1}+...+\sqrt{x_n}-\sqrt{x_{n-1}})=2(\sqrt{x_n}-\sqrt{x_0})$$

2. Apr 26, 2017

Dick

$c_k$ is supposed to be an intermediate value. That is, $\frac{1}{\sqrt{x_{k}}} \le \frac{1}{\sqrt{c_k}} \le \frac{1}{\sqrt{x_{k-1}}}$. Can you check that that works for this choice of $c_k$? Does it work if you change the coefficient?

3. Apr 27, 2017

Karol

$$\frac{1}{\sqrt{x_k}}=\frac{2}{2\sqrt{x_k}}<\frac{2}{\sqrt{x_k}+\sqrt{x_{k-1}}}<\frac{2}{2\sqrt{x_{k-1}}}=\frac{1}{\sqrt{x_{k-1}}}$$
If i change the coefficient 2 into something greater or smaller then each time one side of the equation isn't right.
But there are many values on the graph between $~\frac{1}{\sqrt{x_k}}~$ and $~\frac{1}{\sqrt{x_{k-1}}}$

Last edited: Apr 27, 2017
4. Apr 27, 2017

Dick

Good. So you know $c_k$ is a good choice for an intermediate value. And sure, there are lots of other choices of ways to choose a $c_k$, but this is the one that gives you that nice telescoping sum.

5. Apr 27, 2017

Karol

I myself can't prove that other coefficients fit between $~\frac{1}{\sqrt{x_k}}~$ and $~\frac{1}{\sqrt{x_{k-1}}}~$ but if i look only at $~S_n=2(\sqrt{x_n}-\sqrt{x_0})~$ then anything other than 2 will give the same basic Sn

6. Apr 27, 2017

Dick

I'm not quite sure what you are asking. Yes, anything other than 2 will give you a telescoping sum. But choosing anything other than 2 will not necessarily give you intermediate values. Think what happens if the spacing between $x_k$ and $x_{k-1}$ becomes very small.

7. Apr 27, 2017

Karol

Thank you very much Dick