**Task Overview:**Compute the Riemann sum $\sum_{i=1}^{n} f(x_i) \Delta x$ for $f(x) = x^3$, $[a, b] = [0, 3]$.**Options:**1. $\frac{81}{4} \left( 1 + \frac{1}{n} \right)^2$2. $81 \left( \frac{1}{2} + \frac{1}{n} \right)^2$3. $\left( \frac{9}{2} + \frac{1}{n} \right)^2$4. $\frac{81}{2} \left( 1 + \frac{1}{n} \right) \left( \frac{1}{2} + \frac{1}{n} \right)$5. $\frac{1}{4} \left( 9 + \frac{1}{n} \right)^2$**Explanation:**This task involves computing a Riemann sum for the function $ f(x) = x^3 $ over the interval $[0, 3]$. Riemann sums are used as an approximation of the integral of a function over a specific interval. Each choice represents a possible expression for this computed sum, formed through manipulation of the function and interval given.

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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**Task Overview:**

Compute the Riemann sum $\sum_{i=1}^{n} f(x_i) \Delta x$ for $f(x) = x^3$, $[a, b] = [0, 3]$.

**Options:**

1. $\frac{81}{4} \left( 1 + \frac{1}{n} \right)^2$

2. $81 \left( \frac{1}{2} + \frac{1}{n} \right)^2$

3. $\left( \frac{9}{2} + \frac{1}{n} \right)^2$

4. $\frac{81}{2} \left( 1 + \frac{1}{n} \right) \left( \frac{1}{2} + \frac{1}{n} \right)$

5. $\frac{1}{4} \left( 9 + \frac{1}{n} \right)^2$

**Explanation:**

This task involves computing a Riemann sum for the function $ f(x) = x^3 $ over the interval $[0, 3]$. Riemann sums are used as an approximation of the integral of a function over a specific interval. Each choice represents a possible expression for this computed sum, formed through manipulation of the function and interval given.

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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