The question is: Find the area of the region that lies under the graph of of over the given interval: $f(x) = 4x^3;\;\; 2 \le x \le 5$ My calculation so far:
Width: Change in $x = \frac{(b-a)}n = \frac{5-2}n = \frac3n$
Right Endpoint: $x_k = a+ k\cdot x = 2+k\frac3n = 2+ \frac{3k}n$
Height: \begin{align*}f(x_k) &= f(2+ 3k/n)\\ & = 4(2+3k/n)^3\\ &=4(8 + 36k/n + 6(3k)^2/n^2 + 27k^3/n^3)\\ &= [32+ 144k/n + 216k^2/n^2 + 108k^3/n^3](3/n). \end{align*} I am not sure whether my calculations are right or wrong so far, and then i do not know how to continue after on.
Except at the end where you prematurely multiplied your calculation of $f(x_k)$ by the partition width $3/n$, your calculation is correct so far (this is just $f(x_k)$ - you don't multiply by the width until you calculate the area of the strip).
The next step is to calculate that the area $A_k$ of the $k^{th}$ strip is given by $$A_k = f(x_k)\left(\frac3n\right)$$
Then you sum the areas of all the strips $$A(n) = \sum_{k=1}^n A_k$$ to get an estimate of the area under the curve.
Finally, you let $n \to \infty$ to refine your estimate until it is exact.
And I'll leave you with a warning: you will find that most of the effort you put into calculating $f(x_k)$ (and later $A(n)$) will go to waste in the end, as letting $n \to \infty$ will kill the contributions of most of the terms. But I still recommend going through the exercise, as it will help you to understand what goes on and maybe give a little insight into what was important to know about those other terms.