# How do you use limits to evaluate int 8xdx from [0,2]?

Dec 6, 2016

Evaluating an integral using limits will give you the formula:

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

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

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

#### Explanation:

So for:

${\int}_{0}^{2} 8 x \mathrm{dx}$

$\Delta x = \frac{2}{n}$

${x}_{i} = \frac{2 i}{n}$

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

$= {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \frac{32 i}{n} ^ 2$

We can pull out the constant:

$= {\lim}_{n \to \infty} \frac{32}{n} ^ 2 {\sum}_{i = 1}^{n} i$

$i = \frac{n \left(n + 1\right)}{2}$

$= {\lim}_{n \to \infty} \frac{32}{n} ^ 2 \cdot \frac{n \left(n + 1\right)}{2}$

Now, here you can skip some tedious algebra:

Since the degree of the denominator is 2 and the degree of the numerator is also two, the limit will just be a ratio:

Thus, $= {\lim}_{n \to \infty} \frac{32}{n} ^ 2 \cdot \frac{n \left(n + 1\right)}{2} = 16$