Use the Archimedes-Riemann Theorem to find $\displaystyle \int ^1_0 (x+1)dx $
Proof so far.
So I have $\displaystyle \sum^n_{i=1} i = \frac {(n)(n+1)}{2} $, then how should I proceed? Thanks.
Each subinterval has length:
$\displaystyle {\Delta}x=\frac{b-a}{n}=\frac{1-0}{n}=\frac{1}{n}$
Using the right endpoint method, $\displaystyle x_{k}=a+k{\Delta}x=0+k\cdot\frac{1}{n}=\frac{k}{n}$
Thus, rectangle k has area:
$\displaystyle f(x_{k}){\Delta}x=(1+\frac{k}{n})\cdot \frac{1}{n}$
and the sum of these areas is:
$\displaystyle \sum_{k=1}^{n}f(x_{k}){\Delta}x=\frac{1}{n^{2}}\su m_{k=1}^{n}k+\frac{1}{n}\sum_{k=1}^{n}1$
Knowing the identity: $\displaystyle \sum_{k=1}^{n}k=\frac{n(n+1)}{2}$
we sub that in so we have it all in terms of n.
Getting:
$\displaystyle \frac{1}{n^{2}}\cdot\frac{n(n+1)}{2}+\frac{1}{n}\c dot n$
$\displaystyle =\frac{1}{2n}+\frac{3}{2}$
Now, the more n rectangles we get the closer to the area under the curve,
So, we let $\displaystyle n\to \infty$
$\displaystyle \lim_{n\to {\infty}}\left[\frac{1}{2n}+\frac{3}{2}\right]=\frac{3}{2}$
If you check this against the integral, they should be the same.
This is an important concept to understand when dealing with integration and area. Can you see what is going on?. We are adding up the area of an infinite number of rectangles. As this number of rectangles gets larger and larger, we get the area.