# How do you find the area between the given curve y= x^2 and the x-axis given in the interval [0,1]?

Nov 8, 2015

#### Answer:

The area is $\frac{1}{3}$

#### Explanation:

I am assuming that you do not yet have the Fundamental Theorem of Calculus available to evaluate this, but that you need to evaluate it from a definition.

.${\int}_{a}^{b} {x}^{2} \mathrm{dx} = {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$.

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.)

I prefer to do this type of problem one small step at a time.

${\int}_{0}^{1} {x}^{2} \mathrm{dx}$.

For each $n$, we get

$\Delta x = \frac{b - a}{n} = \frac{1 - 0}{n} = \frac{1}{n}$

And ${x}_{i} = a + i \Delta x = 0 + i \frac{1}{n} = \frac{i}{n}$

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

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

$= {\sum}_{i = 1}^{n} {i}^{2} / {n}^{3}$

$= \frac{1}{n} ^ 3 {\sum}_{i = 1}^{n} {i}^{2}$

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

So,

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

The last thing to do is evaluate the limit as $n \rightarrow \infty$.
I hope it is clear that this amounts to evaluating
${\lim}_{n \rightarrow \infty} \frac{n \left(n + 1\right) \left(2 n + 1\right)}{n} ^ 3$

There are several ways to think about this:

Limit of a Rational Expression

The numerator can be expanded to a plynomial with leading term $2 {n}^{3}$, so the limit as $n \rightarrow \infty$ is $2$.

OR

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

The limit at infinity is $\left(1\right) \left(1\right) \left(2\right) = 2$ as a product of rational expressions.

OR

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

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

So the limit is, again $2$.

However we get it, we get

.${\int}_{0}^{1} {x}^{2} \mathrm{dx} = {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$

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

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

$= \frac{1}{6} \left(2\right)$

$= \frac{1}{3}$