Integral of a area under a straight line as summation

[Karol]

Homework Statement

Homework Equations

Summs:
$$1+2+3+...+n=\frac{n(n+1)}{2}$$
$$1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$

The Attempt at a Solution

$$\Delta x=\frac{b}{n}$$
$$S_n=f\left( \frac{\Delta x}{2} \right)\Delta x+f\left( \Delta x+\frac{\Delta x}{2} \right)\Delta x+...+f\left( (n-1)\Delta x+\frac{\Delta x}{2} \right)\Delta x$$
$$S_n=\left( \frac{b}{2n} \right) \frac{b}{n}+\left( \frac{b}{n}+\frac{b}{2n} \right) \frac{b}{n}+\left( 2\frac{b}{n}+\frac{b}{2n} \right) \frac{b}{n}+...+\left( (n-1)\frac{b}{n}+\frac{b}{2n} \right) \frac{b}{n}$$
$$S_n=\frac{b^2}{2n^2}(1+3+5+...+2n-1)$$
But i was taught, in that chapter, only the two sums from the Relevant Equations

 

[LCKurtz]

Here's a hint to get you a better start. Let $\Delta x_k = x_k - x_{k-1}$ and $c_k = \frac{x_k+x_{k-1}}{2}$ with $x_0=0,~x_n = b$. Then$$
S_n = \sum_{k=1}^n f(c_k)\Delta x_k =\sum_{k=1}^n \frac{(x_k+x_{k-1})}{2}(x_k - x_{k-1})$$Try simplifying that product in the sum and write out a few terms to see what happens. I don't think you will need any of your fancy sum formulas.

 

[vela]

And if you want to continue with your attempt as well, consider the sum $1+2+\cdots+(2n-1)+2n$ and split it up into a sum of the even and odd terms.

 

[Karol]

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

$$S_n=\frac{x_1+x_0}{2}(x_1-x_0)+\frac{x_2+x_1}{2}(x_2-x_1)+\frac{x_3+x_2}{2}(x_3-x_2)+...+\frac{x_n+x_{n-1}}(x_n-x_{n-1})=...=\frac{1}{2}x_n^2$$

 

[LCKurtz]

$$S_n=\frac{x_1+x_0}{2}(x_1-x_0)+\frac{x_2+x_1}{2}(x_2-x_1)+\frac{x_3+x_2}{2}(x_3-x_2)+...+\frac{x_n+x_{n-1}} 2 (x_n-x_{n-1})=...=\frac{1}{2}x_n^2$$

Do you see why it equals $\frac{1}{2}x_n^2$? Did you multiply out the products in the sum as I suggested in post #2? Do you understand how to get the answer, or are you just copying stuff?

 

[Karol]

I know how to get to the answer, i just shortened. but i don't understand the other method suggested:

And if you want to continue with your attempt as well, consider the sum $1+2+\cdots+(2n-1)+2n$ and split it up into a sum of the even and odd terms.

I calculated that $~1+3+5+...+n=n^2$, but not as a split sum of even and odd numbers.