To find left and rigth Riemann sum of \(f(x) = x^{2}\) on the interval [1, 9] with 16 subintervals. The formula for left Riemann sum is,
\(\int_{a}^{b} f(x)dx \approx \Delta x (f(x_0)+f(x_{1})+....f(x_{n-2})+f(x_{n-1})\)

where \(\Delta x = \frac{b-a}{n}\)

Here \(f(x) = x^{2}\) and \([a, b] = [1, 9], n = 16\)

So \(\Delta x = \frac{9-1}{16} = \frac{1}{2}.\)

Now divide the interval [1, 9] into \(n = 16\) subintervals of the length \(\Delta x =\frac{1}{2}.\)

The subintervals are,

\([1, \frac{3}{2}], [\frac{3}{2}, 2], [2, \frac{5}{2}], [\frac{5}{2}, 3], [3, \frac{7}{2}], [\frac{7}{2}, 4]....[8, \frac{17}{2}], [17/2, 9].\)

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,

\(f (x_0) = f (1) = 1^{2} = 1\)

\(f (x_{1}) = f (\frac{3}{2}) = (\frac{3}{2})^{2} = \frac{9}{2}\)

\(f (x_{2}) = f (2) = 2^{2} = 4\)

\(f (x_{3}) = f (\frac{5}{2}) = (\frac{5}{2})^{2} = \frac{25}{5}\)

\(\vdots\)

\(f (x_{14]) = f (8) = 8^{2} = 64\)

\(f (x_{15}) = f (\frac{17}{2}) = (\frac{17}{2})^{2} = \frac{289}{4}\)

Substitute all values in the formula,

\(\int_{1}^{9} x^{2}dx\approx\frac{1}{2}(1+\frac{9}{4}+4+\frac{25}{4}+9+\frac{49}{4}+....+64+\frac{289}{4})\)

\(\approx \frac{1}{2} (1 + 2.25 + 4 + 6.25 + 9 +....+ 64 + 72.25)\)

\(\approx \frac{1}{2} (446)\)

\(\approx 223\)

Therefore left Riemann sum is 223.

To find right Riemann sum.

The formula for right Riemann sum is,

\(\int_{a}^{b} f(x)dx \approx \Delta x(f(x_{1})+f(x_{2})+....f(x_{n-1})+f(x_{n}))\) where

\(\Delta x = \frac{b-a}{n}\)

Here \(f (x) = x^{2} and [a, b] = [1, 9], n = 16.\)

So \(\Delta x = \frac{9-1}{16} = \frac{1}{2}.\)

Now divide the interval [1, 9] into \(n = 16\) subintervals of the length

\(\Delta x =. \frac{1}{2}\)

The subintervals are,

\([1, \frac{3}{2}], [\frac{3}{2}, 2], [2, \frac{5}{2}], [\frac{5}{2}, 3], [3, \frac{7}{2}], [\frac{7}{2}, 4], [4, \frac{9}{2}]....[8, \frac{17}{2}], [\frac{17}{2}, 9].\)

To find right Riemann sum use the right end points of the subintervals.

Here the right endpoints are, \(\frac{3}{2}, 2, \frac{5}{2}, 3...., , \frac{17}{2}, 9.\)

Now evaluate the function at right endpoints,

\(f (x_{1}) = f (\frac{3}{2}) = (\frac{3}{2})^{2} = \frac{9}{4}\)

\(f (x_{2}) = f (2) = 2^{2} = 4\)

\(f (x_{3}) = f (\frac{5}{2}) = (\frac{5}{2})^{2} = \frac{25}{4}\)

\(\vdots\)

\(f (x_{15}) = f (\frac{17}{2}) = (\frac{17}{2})^{2} = \frac{289}{4}\)

\(f (x_{16}) = f (9) = 9^{2} = 81\)

Substitute all values in the formula,

\(\int_{1}^{9} x^{2}dx \approx \frac{1}{2}(\frac{9}{4}+4+\frac{25}{4}+9+\frac{49}{4}+....+\frac{289}{4}+81)\)

\(\approx \frac{1}{2} (2.25 + 4 + 6.25 + 9 +....+ 72.25 + 81)\)

\(\approx \frac{1}{2} (526)\)

\(\approx 263\) Therefore the right Riemann sum is 263.

where \(\Delta x = \frac{b-a}{n}\)

Here \(f(x) = x^{2}\) and \([a, b] = [1, 9], n = 16\)

So \(\Delta x = \frac{9-1}{16} = \frac{1}{2}.\)

Now divide the interval [1, 9] into \(n = 16\) subintervals of the length \(\Delta x =\frac{1}{2}.\)

The subintervals are,

\([1, \frac{3}{2}], [\frac{3}{2}, 2], [2, \frac{5}{2}], [\frac{5}{2}, 3], [3, \frac{7}{2}], [\frac{7}{2}, 4]....[8, \frac{17}{2}], [17/2, 9].\)

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,

\(f (x_0) = f (1) = 1^{2} = 1\)

\(f (x_{1}) = f (\frac{3}{2}) = (\frac{3}{2})^{2} = \frac{9}{2}\)

\(f (x_{2}) = f (2) = 2^{2} = 4\)

\(f (x_{3}) = f (\frac{5}{2}) = (\frac{5}{2})^{2} = \frac{25}{5}\)

\(\vdots\)

\(f (x_{14]) = f (8) = 8^{2} = 64\)

\(f (x_{15}) = f (\frac{17}{2}) = (\frac{17}{2})^{2} = \frac{289}{4}\)

Substitute all values in the formula,

\(\int_{1}^{9} x^{2}dx\approx\frac{1}{2}(1+\frac{9}{4}+4+\frac{25}{4}+9+\frac{49}{4}+....+64+\frac{289}{4})\)

\(\approx \frac{1}{2} (1 + 2.25 + 4 + 6.25 + 9 +....+ 64 + 72.25)\)

\(\approx \frac{1}{2} (446)\)

\(\approx 223\)

Therefore left Riemann sum is 223.

To find right Riemann sum.

The formula for right Riemann sum is,

\(\int_{a}^{b} f(x)dx \approx \Delta x(f(x_{1})+f(x_{2})+....f(x_{n-1})+f(x_{n}))\) where

\(\Delta x = \frac{b-a}{n}\)

Here \(f (x) = x^{2} and [a, b] = [1, 9], n = 16.\)

So \(\Delta x = \frac{9-1}{16} = \frac{1}{2}.\)

Now divide the interval [1, 9] into \(n = 16\) subintervals of the length

\(\Delta x =. \frac{1}{2}\)

The subintervals are,

\([1, \frac{3}{2}], [\frac{3}{2}, 2], [2, \frac{5}{2}], [\frac{5}{2}, 3], [3, \frac{7}{2}], [\frac{7}{2}, 4], [4, \frac{9}{2}]....[8, \frac{17}{2}], [\frac{17}{2}, 9].\)

To find right Riemann sum use the right end points of the subintervals.

Here the right endpoints are, \(\frac{3}{2}, 2, \frac{5}{2}, 3...., , \frac{17}{2}, 9.\)

Now evaluate the function at right endpoints,

\(f (x_{1}) = f (\frac{3}{2}) = (\frac{3}{2})^{2} = \frac{9}{4}\)

\(f (x_{2}) = f (2) = 2^{2} = 4\)

\(f (x_{3}) = f (\frac{5}{2}) = (\frac{5}{2})^{2} = \frac{25}{4}\)

\(\vdots\)

\(f (x_{15}) = f (\frac{17}{2}) = (\frac{17}{2})^{2} = \frac{289}{4}\)

\(f (x_{16}) = f (9) = 9^{2} = 81\)

Substitute all values in the formula,

\(\int_{1}^{9} x^{2}dx \approx \frac{1}{2}(\frac{9}{4}+4+\frac{25}{4}+9+\frac{49}{4}+....+\frac{289}{4}+81)\)

\(\approx \frac{1}{2} (2.25 + 4 + 6.25 + 9 +....+ 72.25 + 81)\)

\(\approx \frac{1}{2} (526)\)

\(\approx 263\) Therefore the right Riemann sum is 263.