Integral Calculus for Riemann Sums Summation of area of square root x from 0-3 I was looking to find the integral from 0 to 3 of f(x) = x^(1/2) for my maths investigation for school. However, I was looking to solve this through a summation of Riemann sums, with a limit of n (the nu,ber of rectangles) approaching infinity. I was able to find an example of how to find the area from 0-1, but I am struggling to be able to use apply this concept to find the area from 0-3.
At the bottom of this page: https://wiki.math.ucr.edu/index.php/Riemann_Sums they solve it from 0-1.
Currently I have a proof that is based off observation that as n approaches infinity, it will reach the exact area. However, I was looking for a mathematical evaluation, and though I understand what they are doing in the article, I'm not sure what part I must do differently to find the area from 0-3 instead. Any help would be greatly appreciated.
Thank you
 A: This is the calculation on the website:
\begin{align}\displaystyle \int_{0}^{1}\sqrt{x}\, dx = \displaystyle \lim_{n\rightarrow\infty}\,\sum_{i=1}^{n}f(x_{i})\cdot\Delta x_{i}\\
\\
=\displaystyle \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{i}{n}\cdot\frac{2i-1}{n^{2}}\\
\\
=\displaystyle \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{2i^{2}-i}{n^{3}}\\
\\
=\displaystyle \lim_{n\rightarrow\infty}\frac{1}{n^{3}}\left(2\cdot\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}\right)\\
\\
=\displaystyle \lim_{n\rightarrow\infty}\left(\frac{2n(n+1)(2n+1)}{6n^{3}}-\frac{n(n+1)}{2n^{3}}\right)\\
\\
=\displaystyle \lim_{n\rightarrow\infty}\,\left(\frac{4n^{3}}{6n^{3}}+\frac{n^{2}}{2n^{3}}\right)\qquad\qquad(\textrm{for~large~}n)\\
\\
=\displaystyle \frac{2}{3}+0\\
\\
=\displaystyle \frac{2}{3}.
\end{align}
$$$$
They are doing calculations with rectangles of endpoints $\ x_i\ ;\ $ the left-most rectangle has left endpoint $\ x_0=0\ $ and the right-most rectangle has right endpoint at $\ x_n=1.$ Therefore, all you need to do is use rectangle three times the width as in their example, and you're done!
But you must be careful, because you're not just getting all of the rectangles in their example and making them three times wider. For example, your left-most rectangle has left endpoint $\ x_0=0\ $ and right endpoint $\ x_1=\frac{3}{n^2},\ $ the next rectangle has left endpoint  $\ x_1=\frac{3}{n^2},\ $ and right endpoint $\ x_2=3 \times \frac{4}{n^2},\ $ etc. So you're using one third the amount of rectangles as they are in their calculation, and these rectangles are three times the width as the ones in the example.
Now we must replace $\ x_i\ $ with $\ 3x_i\ $ which means you also must also replace $\ f(x_i)\ $ with $\ f(3x_i),\ $ which equals $\ \large{\sqrt{3 \frac{i^2}{n^2} } = \sqrt{3}\ \frac{i}{n}}.$
Comparing with their example, this gives us:
\begin{align}
\\
\displaystyle \int_{0}^{3}\sqrt{x}\, dx = \displaystyle \lim_{n\rightarrow\infty}\,\sum_{i=1}^{n}f(3x_{i})\cdot\ 3 \Delta x_{i}\\
\\
\displaystyle \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\sqrt{3}\frac{i}{n}\cdot3 \Delta x_{i}\\
\\
=3\sqrt{3}\ \displaystyle \lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x_{i})\cdot\Delta x_{i}\qquad (!)\\
\\
=3\sqrt{3}\cdot \frac{2}{3} \qquad \text{using the result from their example}\\
\\
=2\sqrt{3}.
\end{align}
