# How do you use limits to evaluate int(x^5+x^2)dx from [0,2]?

Jan 6, 2017

Please see the explanation section below.

#### Explanation:

I assume that you are given ${\sum}_{i = 1}^{n} {i}^{5} = \frac{{n}^{2} \left(2 {n}^{2} + 2 n - 1\right) {\left(n + 1\right)}^{2}}{12}$ and ${\sum}_{i = 1}^{n} {i}^{2} = \frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}$.

${\int}_{0}^{2} \left({x}^{5} + {x}^{2}\right) \mathrm{dx}$

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

$\Delta x = \frac{b - a}{n}$, in this case $\Delta x = \frac{2}{n}$

${x}_{i} = a + i \Delta x$, in this case ${x}_{i} = i \frac{2}{n}$.

Since $f \left(x\right) = {x}^{5} + {x}^{2}$, we have

${\int}_{0}^{2} \left({x}^{5} + {x}^{2}\right) \mathrm{dx} = {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \left({i}^{5} \frac{32}{n} ^ 5 + {i}^{2} \frac{4}{n} ^ 2\right) \frac{2}{n}$

$= {\lim}_{n \rightarrow \infty} \left(\frac{64}{n} ^ 6 {\sum}_{i = 1}^{n} {i}^{5} + \frac{8}{n} ^ 3 {\sum}_{i = 1}^{n} {i}^{2}\right)$

$= {\lim}_{n \rightarrow \infty} \left(\frac{64}{n} ^ 6 \left(\frac{{n}^{2} \left(2 {n}^{2} + 2 n - 1\right) {\left(n + 1\right)}^{2}}{12}\right) + \frac{8}{n} ^ 3 \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right)\right)$

$= {\lim}_{n \rightarrow \infty} \left(\frac{64}{12} \left(\frac{{n}^{2} \left(2 {n}^{2} + 2 n - 1\right) {\left(n + 1\right)}^{2}}{n} ^ 6\right) + \frac{8}{6} \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{n} ^ 3\right)\right)$

$= \frac{64}{12} \left(2\right) + \frac{8}{6} \left(2\right) = \frac{32}{3} + \frac{8}{3} = \frac{40}{3}$