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(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.

Question
Confidence intervals
asked 2021-01-31
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.

Answers (1)

2021-02-01
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.
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