Hey Synx:
These can be precarious if you don't see what's going on, but they help in showing how integration works by adding up the rectangles.
Let's do the right hand side and then you try the left hand side. You should get the same answer.
\(\displaystyle \int_{-1}^{3}{-x^{2}+2x+3}dx\), using a Riemann sum with n intervals.
Each subinterval will have length:
\(\displaystyle \Delta{x}=\frac{b-a}{n}=\frac{3-(-1)}{n}=\frac{4}{n}\)
The subintervals are divided into points:
\(\displaystyle x_{k}=a+k\Delta{x}=-1+\frac{4k}{n}\)
Thus, rectangle k has area:
\(\displaystyle \L\\f(x_{k})\Delta{x}=[-(-1+\frac{4k}{n})^{2}+2(-1+\frac{4k}{n})+3](\frac{4}{n})\)
This simplifies to(you can do the algebra):
\(\displaystyle \L\\\frac{-64k(k-n)}{n^{3}}\)
Sum up these areas:
\(\displaystyle \L\\\sum_{k=1}^{n}f(x)\Delta{x}=\frac{-64}{n^{3}}\sum_{k=1}^{n}k^{2}+\frac{64}{n^{2}}\sum_{k=1}^{n}k\)
The sum of the \(\displaystyle k^{2}\) is \(\displaystyle \frac{n(n+1)(2n+1)}{6}\)
The sum of \(\displaystyle k\) is \(\displaystyle \frac{n(n+1)}{2}\)
Use those in the summation(more algebra):
\(\displaystyle \L\\\frac{-64}{n^{3}}\left(\frac{n(n+1)(2n+1)}{6}\right)+\frac{64}{n^{2}}\left(\frac{n(n+1)}{2}\right)\)
Which gives(more algebra):
\(\displaystyle \L\\\lim_{n\to\infty}\left[\frac{-32(n+1)(2n+1)}{3n^{2}}+\frac{32(n+1)}{n}\right]\)
Expand out and get:
\(\displaystyle \L\\\lim_{n\to\infty}\left(\frac{32}{3}-\frac{32}{3n^{2}}\right)\)
Now, see the limit. It's the same as if you'd performed the integration, only a lot more work. This is what integration does. It sums up all the rectangles. Pretty cool, huh?. It was some mighty beautiful minds that figured this all out.
Lots of algebra and knowing the summation identities. Try to factor n out of the sum leaving only the summation in terms of k.
For left-hand method, you use \(\displaystyle a+(k-1)\Delta{x}\)
