Riemann sums are a good illustration of what intgeration is. The summation of the n rectangles to arrive at the area under a curve.
\(\displaystyle \L\\f(x)=\frac{x^{2}}{3}+7\)
If we divide the interval into n equal parts, each will have length \(\displaystyle \L\\{\Delta}x=\frac{2-0}{n}\)
Using the right endpoint method, the right endpoint of each subinterval is:
\(\displaystyle \L\\x_{k}=a+k{\Delta}x=0+\frac{2k}{n}\)
The kth rectangle has area: \(\displaystyle \L\\f(x_{k}){\Delta}x=\left(\frac{\left(\frac{2k}{n}\right)^{2}}{3}+7\right)\frac{2}{n}=\frac{8k^{2}}{3n^{3}}+\frac{14}{n}\)
Now the sum of the rectangles areas is:
\(\displaystyle \L\\\sum_{k=1}^{n}f(x_{k}){\Delta}=\frac{8}{3n^{3}}\sum_{k=1}^{n}k^{2}+\frac{14}{n}\sum_{k=1}^{n}1\)
Remember, the sum of the squares is \(\displaystyle \L\\k^{2}=\frac{n(n+1)(2n+1)}{6}\)
So, you have:
\(\displaystyle \L\\\frac{8}{3n^{3}}\cdot\frac{n(n+1)(2n+1)}{6}+14\)
Now, exapnd and take the limit as n approaches infinity:
\(\displaystyle \L\\\lim_{n\to\infty}(\frac{4}{3n}+\frac{4}{9n^{2}}+\frac{134}{9})\)
It ain't difficult to see the limit now, is it?.
Do you get the same thing as if you integrate the 'easy' way?.
