# How do you use limits to evaluate int x^2dx from [0,4]?

Feb 5, 2017

#### Explanation:

Here is a limit definition of the definite integral. (I hope it's the one you are using.) I will use what I think is somewhat standard notation in US textbooks.

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

Where, for each positive integer $n$, we let $\Delta x = \frac{b - a}{n}$

And for $i = 1 , 2 , 3 , . . . , n$, we let ${x}_{i} = a + i \Delta x$. (These ${x}_{i}$ are the right endpoints of the subintervals.)
We'll do one small step at a time.

${\int}_{0}^{4} {x}^{2} \mathrm{dx}$.

Find $\Delta x$

For each $n$, we get

$\Delta x = \frac{b - a}{n} = \frac{4 - 0}{n} = \frac{4}{n}$

Find ${x}_{i}$

And ${x}_{i} = a + i \Delta x = 0 + i \frac{4}{n} = \frac{4 i}{n}$

Find $f \left({x}_{i}\right)$

$f \left({x}_{i}\right) = {\left.{x}_{i}\right.}^{2} = {\left(\frac{4 i}{n}\right)}^{2}$

$= \frac{16 {i}^{2}}{n} ^ 2$

Find and simplify ${\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$ in order to evaluate the sum.

${\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta t = {\sum}_{i = 1}^{n} \left(\frac{16 {i}^{2}}{n} ^ 2\right) \frac{4}{n}$

$= {\sum}_{i = 1}^{n} \left(\frac{64 {i}^{2}}{n} ^ 3\right)$

$= \frac{64}{n} ^ 2 {\sum}_{i = 1}^{n} \left({i}^{2}\right)$

Evaluate the sum

$= \frac{64}{n} ^ 3 \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right)$

(We used a summation formula in the previous step.)

Rewrite before finding the limit

${\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x = \frac{64}{6} \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{n} ^ 3\right)$

$= \frac{32}{3} \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{n} ^ 3\right)$

Now we need to evaluate the limit as $n \rightarrow \infty$.

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

$= \left(1\right) \left(1\right) \left(2\right) = 2$

To finish the calculation, we have

${\int}_{0}^{4} {x}^{2} \mathrm{dx} = {\lim}_{n \rightarrow \infty} \left(\frac{32}{3} \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{n} ^ 3\right)\right)$

$= \frac{32}{3} \left(2\right) = \frac{64}{3}$