I have to calculate the Right Hand Sum of an integral.
$$f(x)=\frac x2 \ \ [1,4]$$
I am wondering if the procedure is done right. First process I will do is rewrite the problem into an integral: \begin{equation}\int_1^4 f(x)\ dx=\int_1^4 \frac x2 \ \ dx \end{equation}
The integral evaluates to the following $\frac{15}{4}$ Knowing that the right answer is that I proceeded to calculated the right hand sum.
\begin{align}\triangle x&=\frac{b-a}{n}=\frac{4-1}{n}=\frac 3n\\ x^*&=a+k\triangle x=1+\frac{3}{n}k \end{align}
From here I proceeded to the following:
\begin{align}\lim_\limits{n\to \infty}\sum_{k=1}^n f(x^*)\triangle x \\ \lim_\limits{n\to \infty}\sum_{k=1}^n \frac{(1+\frac{3k}{n})}{2} \frac{3}n \end{align}
At the end of this I get the answer being $\frac{15}4$. Is the procedure right? I just find it weird my answer is really close to me regularly integrating it I thought there would be some error in it?
Best Answer
Indeed we have $\sum_{k=1}^{n}\frac{1+\frac{3k}{n}}{2}\frac{3}{n}=\frac{3}{2n}(\sum_{k=1}^{n}1+\frac{3}{n}\sum_{k=1}^{n}k)=\frac{3}{2n}(n+\frac{3}{n}(1+..+n))=\frac{3}{2}+\frac{9}{2n^{2}}\frac{n(n+1)}{2}=\frac{3}{2}+\frac{9}{4}+\frac{9}{4n}=\frac{15}{4}+\frac{9}{4n}$
and taking the limit we obtain the required result.
This is to be expected since for the limiting value, if it exists, the Riemann sum is defined as the definite Riemann integral of the function over the domain,
$$\int_{a}^{b}f(x)dx=\lim_{||\Delta x||\rightarrow0}\sum_{i=1}^{n}f(x_{i}*)\Delta x_{i}.$$
Indeed we have $\int_{1}^{4}\frac{x}{2}dx=\frac{15}{4}$ as expected, and the left hand sum $\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}\frac{1+\frac{3k}{n}}{2}\frac{3}{n}$ also converges to the same value. (For a one-dimensional Riemann sum over domain $[a,b]$, as the maximum size of a partition element shrinks to zero (that is the limit of the norm of the partition goes to zero), some functions will have all Riemann sums converge to the same value)