So, I was asked to do this integral using the limit method (or the Riemann Sum)
$$\int_0^3 x{\sqrt {3-x}}\;dx $$
And, I do it like this:
$$\int_0^3 x{\sqrt {3-x}}\;dx $$
Firstly, I determine the $\Delta x$ and $c_i$
$$\Delta x = \frac{b-a}{n}$$
$$\Delta x = \frac{3-0}{n}$$
$$\Delta x = \frac{3}{n}$$
Using the right end point :
$$c_i=a+\Delta xi$$
$$c_i=0+\frac{3}{n}i$$
$$c_i=\frac{3i}{n}$$
Then, evaluating the Integral using limit:
$$\lim_{x\to\infty}\;\sum_{i=1}^n\;f(c_i)\Delta x $$$$\lim_{n\to\infty}\;\sum_{i=1}^n\;f(c_i)\Delta x $$
$$\lim_{x\to\infty}\;\sum_{i=1}^n\;f\left(\frac{3i}{n}\right)\left(\frac{3}{n}\right)$$$$\lim_{n\to\infty}\;\sum_{i=1}^n\;f\left(\frac{3i}{n}\right)\left(\frac{3}{n}\right)$$
$$\lim_{x\to\infty}\;\sum_{i=1}^n\;\left(\frac{3i}{n}{\sqrt {3-\frac{3i}{n}}}\right)\;\left(\frac{3}{n}\right)$$$$\lim_{n\to\infty}\;\sum_{i=1}^n\;\left(\frac{3i}{n}{\sqrt {3-\frac{3i}{n}}}\right)\;\left(\frac{3}{n}\right)$$
$$\lim_{x\to\infty}\;\left(\frac{9 \sqrt {3}}{n^\frac {5}{2}}\right)\;\sum_{i=1}^n\;i{\sqrt {n-i}}$$$$\lim_{n\to\infty}\;\left(\frac{9 \sqrt {3}}{n^\frac {5}{2}}\right)\;\sum_{i=1}^n\;i{\sqrt {n-i}}$$
And, now I'm stuck here. Is there a way to do a summation that has a square root in it?
notes : if you find error in my calculation please let me know