## anonymous one year ago Need help with Riemann sum equation

1. anonymous

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

2. SolomonZelman

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: $$[1,~~~1+5/3]$$ ---> $$[1,~~~8/3]$$ $$[8/3,~~~8/3+5/3]$$ ---> $$[8/3,~~~13/3]$$ $$[13/3,~~~13/3+5/3]$$ ---> $$[13/3,~~~6]$$

3. SolomonZelman

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)*.

4. SolomonZelman

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