Consider the function f(x) = x^{2} on the interval [1, 9]. Let P be a uniform partition of [1,9] with 16 sub-intervals. Compute the left and right Riemann sum of f on the partition. Use exact values.

Consider the function $f\left(x\right)={x}^{2}$ on the interval [1, 9]. Let P be a uniform partition of [1,9] with 16 sub-intervals. Compute the left and right Riemann sum of f on the partition. Use exact values.
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To find left and rigth Riemann sum of $f\left(x\right)={x}^{2}$ on the interval [1, 9] with 16 subintervals. The formula for left Riemann sum is, ${\int }_{a}^{b}f\left(x\right)dx\approx \mathrm{\Delta }x\left(f\left({x}_{0}\right)+f\left({x}_{1}\right)+....f\left({x}_{n-2}\right)+f\left({x}_{n-1}\right)$
where $\mathrm{\Delta }x=\frac{b-a}{n}$
Here $f\left(x\right)={x}^{2}$ and $\left[a,b\right]=\left[1,9\right],n=16$
So $\mathrm{\Delta }x=\frac{9-1}{16}=\frac{1}{2}.$
Now divide the interval [1, 9] into $n=16$ subintervals of the length $\mathrm{\Delta }x=\frac{1}{2}.$
The subintervals are,
$\left[1,\frac{3}{2}\right],\left[\frac{3}{2},2\right],\left[2,\frac{5}{2}\right],\left[\frac{5}{2},3\right],\left[3,\frac{7}{2}\right],\left[\frac{7}{2},4\right]....\left[8,\frac{17}{2}\right],\left[17/2,9\right].$
To find left Riemann sum use the left end points of the subintervals.
Here the left endpoints are, 1, $\frac{3}{2},2,....,8,\frac{17}{2}.$
Now evaluate the function at left endpoints,
$f\left({x}_{0}\right)=f\left(1\right)={1}^{2}=1$
$f\left({x}_{1}\right)=f\left(\frac{3}{2}\right)=\left(\frac{3}{2}{\right)}^{2}=\frac{9}{2}$
$f\left({x}_{2}\right)=f\left(2\right)={2}^{2}=4$
$f\left({x}_{3}\right)=f\left(\frac{5}{2}\right)=\left(\frac{5}{2}{\right)}^{2}=\frac{25}{5}$
$⋮$
$f\left({x}_{14}\right)=f\left(8\right)={8}^{2}=64$
$f\left({x}_{15}\right)=f\left(\frac{17}{2}\right)=\left(\frac{17}{2}{\right)}^{2}=\frac{289}{4}$
Substitute all values in the formula,
${\int }_{1}^{9}{x}^{2}dx\approx \frac{1}{2}\left(1+\frac{9}{4}+4+\frac{25}{4}+9+\frac{49}{4}+....+64+\frac{289}{4}\right)$
$\approx \frac{1}{2}\left(1+2.25+4+6.25+9+....+64+72.25\right)$
$\approx \frac{1}{2}\left(446\right)$
$\approx 223$
Therefore left Riemann sum is 223.
To find right Riemann sum.
The formula for right Riemann sum is,
${\int }_{a}^{b}f\left(x\right)dx\approx \mathrm{\Delta }x\left(f\left({x}_{1}\right)+f\left({x}_{2}\right)+....f\left({x}_{n-1}\right)+f\left({x}_{n}\right)\right)$ where
$\mathrm{\Delta }x=\frac{b-a}{n}$
Here $f\left(x\right)={x}^{2}and\left[a,b\right]=\left[1,9\right],n=16.$
So $\mathrm{\Delta }x=\frac{9-1}{16}=\frac{1}{2}.$
Now divide the interval [1, 9] into $n=16$ subintervals of the length
$\mathrm{\Delta }x=.\frac{1}{2}$
The subintervals are,