Question

How do you find the Riemann sum for $f(x) = {x^2} + 3x$ over the interval $[0,8]$?

Answer 1

Hint:
In mathematics, Riemann sum is defined as a certain kind of approximation of an integral by a finite sum. In this question, we need to find the Riemann sum for $f(x) = {x^2} + 3x$ over the interval $[0,8]$ for which we will use the formula, $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f({x_i})\Delta x = \int\limits_a^b {f(x)dx} } $

Complete step by step solution:
According to the given information, we have to evaluate the Riemann sum for $f(x) = {x^2} + 3x$.
The formula we are using to do so looks like, $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f({x_i})\Delta x = \int\limits_a^b {f(x)dx} } $
Thus, when we put all the values, the expression becomes
\[\int\limits_0^8 {({x^2} + 3x)dx} \]
We need to use this information to plug in values into our Riemann sum formula.
$\Delta x = \dfrac{{b - a}}{n}$
${x_i} = a + i\Delta x$
Thus,
$\Delta x = \dfrac{{8 - 0}}{n} = \dfrac{8}{n}$
And on further evaluation we get,
${x_i} = \dfrac{{8i}}{n}$
So, in order to get a Riemann sum:
 \[\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\dfrac{8}{n}[{{(\dfrac{{8i}}{n})}^2} + 3(\dfrac{{8i}}{n})]} \]
\[ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\dfrac{8}{n}[(\dfrac{{64}}{{{n^2}}}){i^2} + (\dfrac{{24}}{n})i]} \]
We need to note that since $i$ is our variable and $n$ is our constant, we can pull those to the front.
 The expression then becomes,
 \[ \Rightarrow \mathop {\lim }\limits_{n \to \infty } [\dfrac{{512}}{{{n^3}}}\sum\limits_{i = 1}^n {{i^2} + \dfrac{{192}}{{{n^2}}}\sum\limits_{i = 1}^n i ]} \]
 Recall from summation formulas:
 \[\sum\limits_{i = 1}^n i = \dfrac{{n(n + 1)}}{2}\]
 And, \[\sum\limits_{i = 1}^n {{i^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}} \]
 On substituting the obtained values, we get
 \[ \Rightarrow \mathop {\lim }\limits_{n \to \infty } [\dfrac{{512}}{{{n^3}}}\sum\limits_{i = 1}^n {\dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{{192}}{{{n^2}}}\sum\limits_{i = 1}^n {\dfrac{{n(n + 1)}}{2}} ]} \]
 Now, the degree of the denominators are the same as the numerators, so it will result in the sum of two ratios.

\[ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{800}}{3} \approx 266.7\]

Note:
A Riemann sum is defined as a certain kind of approximation of an integral by a finite sum. The formula used to evaluate the Riemann sum is, $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f({x_i})\Delta x = \int\limits_a^b {f(x)dx} } $.