Evaluate and interpret the following Riemann sums for ƒ(x) = 1 - x2 on the interval [a, b] with n equally spaced subintervals.A midpoint Riemann sum with [a, b] = [1, 3] and n = 4

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Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Evaluate and interpret the following Riemann sums for ƒ(x) = 1 - x² on the interval [a, b] with n equally spaced subintervals.
A midpoint Riemann sum with [a, b] = [1, 3] and n = 4

Expert Solution

Riemann sum for the function.

It is given that the function $f (x) = 1 - x^{2}$ on the interval [a, b].

The interval $[a, b]$ is divided into $n$ equally spaced subintervals defined by the points $a = x_{0} < x_{1} < . . . < x_{n - 1} < x_{n} = b$ where $x_{0} = a, x_{1} = a + \frac{b - a}{n}, x_{2} = a + \frac{2 (b - a)}{n} . . . ., x_{n - 1} = a + \frac{(n - 1) (b - a)}{n} a n d x_{n - 1} = b$ .

Here, $|x_{i + 1} - x_{i}| = \frac{b - a}{n}$ .

Thus, the subintervals are $[x_{0}, x_{1}], [x_{1}, x_{2}], [x_{3}, x_{4}], . . ., [x_{n - 1}, x_{n}]$ and width of each subinterval is $|x_{i + 1} - x_{i}| = \frac{b - a}{n}$ .

The Riemann sum of the function is

$R = \sum_{i = 1}^{n} f (t_{i}) |x_{i} - x_{i - 1}| w h e r e t_{i} i s a n y n u m b e r i n [x_{i - 1}, x_{i}]$

since every subinterval are equal and $f (x) = 1 - x^{2}$ .

$R = \sum_{i = 1}^{n} (1 - {(t_{i})}^{2}) (\frac{b - a}{n}) R = (\frac{b - a}{n}) \sum_{i = 1}^{n} (1 - {(t_{i})}^{2}) R = (\frac{b - a}{n}) [1 - {(t_{1})}^{2} + 1 - {(t_{2})}^{2} + . . . + 1 - {(t_{n})}^{2}] R = \sum_{i = 1}^{n} (1 - {(t_{i})}^{2}) (\frac{b - a}{n}) R = (\frac{b - a}{n}) \sum_{i = 1}^{n} (1 - {(t_{i})}^{2}) R = (\frac{b - a}{n}) [n - [{(t_{1})}^{2} + {(t_{2})}^{2} + . . . + {(t_{n})}^{2}]] w h e r e t_{i} \in [x_{i - 1}, x_{i}]$

Riemann sum for the function is $R = (\frac{b - a}{n}) [n - [{(t_{1})}^{2} + {(t_{2})}^{2} + . . . + {(t_{n})}^{2}]] w h e r e t_{i} \in [x_{i - 1}, x_{i}]$ .

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