Question

Need help with Riemann sum equation

Answer 1

\[\int\limits_{1}^{6}4x-2dx\]
n=3, so using 3 rectangles and left endpoints

Answer 2

If you want the rectangles to be even,

\(\Delta x=\dfrac{b-a}{n}=\dfrac{6-1}{3}=\dfrac{5}{3}\)

So your intervals
\([x_0,~~~x_1]\)
\([x_1,~~~x_2]\)
\([x_2,~~~x_3]\)

(And \([x_1=x_0+\Delta x]\) )

are going to be:
\([1,~~~1+5/3]\) ---> \([1,~~~8/3]\)
\([8/3,~~~8/3+5/3]\) ---> \([8/3,~~~13/3]\)
\([13/3,~~~13/3+5/3]\) ---> \([13/3,~~~6]\)

Answer 3

You are doing *left* endpoints, and therefore your first rectangle will have the height of *f(1)*. Then, the second rectangle will have a height of *f(8/3)* and the third rectangle will have a height of *f(13/3)*.

Answer 4

So basically, your area by reinman sums:

assuming that \(\Delta x\) is same (which I found to be 5/3)
for all intervals (as I supposed in the beginning),

*AND* with the appropriate heights (which I listed accordingly)

\(\large\color{black}{ \displaystyle {\rm A}=\frac{5}{3}f(1)+\frac{5}{3}f\left(\frac{8}{3}\right)+\frac{5}{3}f\left(\frac{13}{3}\right) }\)

Adding three rectangles. \(\Uparrow\)

And then after factoring you get:

\(\large\color{black}{ \displaystyle {\rm A}=\frac{5}{3}\color{red}{\left\{\color{black}{f(1)+f\left(\frac{8}{3}\right)+f\left(\frac{13}{3}\right)}\right\}} }\)