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

Dec 7, 2016

The formula looks like:

${\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x = {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

${\int}_{0}^{8} \left({x}^{2} + 3 x\right) \mathrm{dx}$

We can use this information to plug in values into our Riemann sum formula.

$\Delta x = \frac{b - a}{n}$

${x}_{i} = a + i \Delta x$

Therefore:

$\Delta x = \frac{8 - 0}{n} = \frac{8}{n}$

${x}_{i} = 0 + i \left(\frac{8}{n}\right) = \frac{8 i}{n}$

So, as a Riemann sum:

${\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \left(\frac{8}{n}\right) \left[{\left(\frac{8 i}{n}\right)}^{2} + 3 \left(\frac{8 i}{n}\right)\right]$

$= {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \left(\frac{8}{n}\right) \left[\left(\frac{64}{n} ^ 2\right) {i}^{2} + \left(\frac{24}{n}\right) i\right]$

Note: Since $i$ is our variable and $n$ is our constant, we can pull those to the front.

${\lim}_{n \to \infty} \left[\frac{512}{n} ^ 3 {\sum}_{i = 1}^{n} {i}^{2} + \frac{192}{n} ^ 2 {\sum}_{i = 1}^{n} i\right]$

Recall from summation formulas:

${\sum}_{i = 1}^{n} i = \left(\frac{n \left(n + 1\right)}{2}\right)$
${\sum}_{i = 1}^{n} {i}^{2} = \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right)$

So, we'll have an awesome looking function:

${\lim}_{n \to \infty} \left[\frac{512}{n} ^ 3 \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right) + \frac{192}{n} ^ 2 \left(\frac{n \left(n + 1\right)}{2}\right)\right]$

Now, since the degree of the denominators are the same as the numerators, it will result in the sum of two ratios.

$= {\lim}_{n \to \infty} \left[\frac{512 \cdot 2}{6} + \frac{192}{2}\right] = \frac{800}{3} \approx 266.7$