nCompute the Riemann sum > f(x;)Ax for f(x) = x³, [a, b] = [0, 3].i=18111+4n211812n(है(+) ( + )9.1n81112n219 +O 14n+

Given: fx=x3,a,b=0,3 We want to calculate ∑i=1nfxi∆x

Answered: Compute the Riemann sum E f(x;)Ax for…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

n
Compute the Riemann sum > f(x;)Ax for f(x) = x³, [a, b] = [0, 3].
i=1
81
1
1+
4
n
2
1
1
81
2
n
(है
(+) ( + )
9.
1
n
81
1
1
2
n
2
1
9 +
O 1
4
n
+

Expert Solution

Step 1

Given:

$f (x) = x^{3}, [a, b] = [0, 3]$

We want to calculate $\sum_{i = 1}^{n} f (x_{i}) ∆ x$

Step 2

Calculation:

Given $f (x) = x^{3}, [a, b] = [0, 3]$

Therefore

$∆ x = \frac{b - a}{n} a n d x_{i} = a + i ∆ x$

Hence

$∆ x = \frac{3 - 0}{n} \Rightarrow ∆ x = \frac{3}{n}$

Now

$x_{i} = 0 + i (\frac{3}{n}) \Rightarrow x_{i} = \frac{3 i}{n}$

Therefore Riemann sum

$\begin{array}{rcl} \sum_{i = 1}^{n} f (x_{i}) ∆ x & = & \sum_{i = 1}^{n} {(\frac{3 i}{n})}^{3} (\frac{3}{n}) \\ = & \frac{3}{n} \sum_{i = 1}^{n} \frac{27 i^{3}}{n^{3}} \\ = & \frac{81}{n^{4}} \sum_{i = 1}^{n} i^{3} \\ = & \frac{81}{n^{4}} \times \frac{n^{2} {(n + 1)}^{2}}{4} \\ . . . . . . . . . . \{\sum_{i = 1}^{n} i^{3} = \frac{n^{2} {(n + 1)}^{2}}{4}\} \\ = & \frac{81}{4} \times \frac{{(n + 1)}^{2}}{n^{2}} \\ = & \frac{81}{4} {(\frac{n + 1}{n})}^{2} \\ \sum_{i = 1}^{n} f (x_{i}) ∆ x & = & \frac{81}{4} {(1 + \frac{1}{n})}^{2} \end{array}$

Hence $\begin{array}{rcl} \sum_{i = 1}^{n} \end{array} \begin{array}{rcl} f \end{array} \begin{array}{rcl} (x_{i}) \end{array} \begin{array}{rcl} ∆ \end{array} \begin{array}{rcl} x \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} \frac{81}{4} \end{array} {\begin{array}{rcl} (1 + \frac{1}{n}) \end{array}}^{2}$

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