Use Riemann Sums to Find Area bounded by Curve How can I find the area bound by $\;x=0,\, x=1,\;$ the $\;x$-axis ($y = 0$) and $\;y=x^2+2x\;$ using Riemann sums? 
I want to use the right-hand sum. Haven't really found any good resources online to explain the estimation of areas bounded by curves, hoping anyone here can help?
By the way, I would like there to be 100 intervals.

Right Riemann Sums place the right corner of the rectangles on the curve.
Right Riemann Sums are an overestimation of area because of all the extra space that is not under the curve that is still calculated in the area because it is inside the rectangles.
For your problem, we have:
$f(x) = x^{2} + 2x$
With: $a = 0, b = 1$ and $100$ rectangles, we have $\frac{b-a}{n} = \frac{1-0}{100} = \frac{1}{100}$
Our formula now becomes:
$\frac{b-a}{n} [f(0) + f(\frac{1}{100}) + f(\frac{2}{100})+ \cdots + f(\frac{100}{100})]$
$\bullet$ For Left Riemann, plug in all numbers except last.
$\bullet$ For Right Riemann, plug in all numbers except first.
$\bullet$ For Midpoint Riemann, average the numbers and plug those in.
So, using Right Riemann, we have:
$f(\frac{1}{100}) = \frac{1^{2}}{100^{2}} + \frac{2}{100} = \frac{201}{10000}$
$f(\frac{2}{100}) = \frac{2^2}{100^{2}} + \frac{4}{100} = \frac{101}{2500}$
$\cdots$
$f(\frac{100}{100}) = \frac{100^2}{100^{2}} + \frac{200}{100} = 3$
You'll get this set of values using WA.
Now, using the formula above, we have:
$\frac{1}{100}[f(\frac{1}{100}) + f(\frac{2}{100})+ \cdots + f(\frac{100}{100})] = (\frac{1}{100})(\frac{26967}{200}) = 1.34835$
Comparing that with the actual $\int_0^1 (x^{2} + 2x) dx = \frac{4}{3} = 1.33333$, we see, as expected, that we got an overestimate.
You can compare these two with the nice pointer by amWhy at Wolfram Math World.
You should repeat this for left and midpoint and compare so you understand!
Regards
