# How do you use limits to find the area between the curve y=x^4 and the x axis from [0,5]?

May 26, 2017

$625$

#### Explanation:

It is better to integrate than to use limits in the problem:

${\int}_{0}^{5} \left[{x}^{4}\right] \mathrm{dx} = {x}^{5} / 5 = {5}^{4} = 625$

May 26, 2017

$625$

#### Explanation:

With limits you say... Ok.

Imaging we have some function $f \left(x\right)$ and we want to find the area underneath it in the interval $\left[a , b\right]$. We first start by dividing the area into tiny rectangles like so: This means the height of each rectangle is equal to the value of the function at that point. Refer to ${R}_{1}$. The height of ${R}_{1}$ is equal to $f \left({x}_{0}\right)$, and the area of it is equal to $f \left({x}_{0}\right) \Delta x$ where $\Delta x$ is just the constant width of each rectangle.

Therefore the area under $\left[a , b\right]$ is in the diagram above is:
${\sum}_{n = 0}^{N} {R}_{n} = {\sum}_{n = 0}^{N} f \left(a + n \Delta x\right) \Delta x = \Delta x {\sum}_{n = 0}^{N} f \left(a + n \Delta x\right)$
Where $N = \frac{| a - b |}{\Delta x}$

We would also say that for smaller and smaller values of $\Delta x$, our approximation of the area will become better and better. Sound familiar? Let's use a limit!

$A r e a = {\lim}_{\Delta x \to 0} {\sum}_{n = 0}^{N} f \left(a + n \Delta x\right) \Delta x$

Now that is some ugly stuff. Let's simplify this a bit and that advantage that the interval you want to find is $\left[0 , 5\right]$ which starts at $0$ obviously:

$A r e a = {\lim}_{\Delta x \to 0} {\sum}_{n = 0}^{N} f \left(n \Delta x\right) \Delta x$

Another way of describing $\Delta x$ is $\frac{| a - b |}{N}$ (see above equations). So, in this case, it's $\frac{| a - b |}{N}$. Since we are applying the limit, there's no need for this to be intuitive and we can go ahead and say: ${\lim}_{N \to \infty}$
Finally:
$A r e a = {\lim}_{N \to \infty} {\sum}_{n = 1}^{N} f \left(\frac{5}{N} n\right) \frac{5}{N}$

Oh, did I mention that $n = 0$ and $n = 1$ won't matter, but let's make it $1$ for the sake of calculations. And there we have it folks, our very own limit for an area.

To do anything with it, we need to expand stuff. Starting with $f \left(x\right)$ and then factoring out any things that can be factored:

${\lim}_{N \to \infty} {\sum}_{n = 1}^{N} f \left(\frac{5}{N} n\right) \frac{5}{N} = {\lim}_{N \to \infty} \frac{5}{N} {\sum}_{n = 1}^{N} {\left(\frac{5}{N}\right)}^{4} {n}^{4}$
$= {\lim}_{N \to \infty} {5}^{5} / {N}^{5} {\sum}_{n = 1}^{N} {n}^{4}$

And we need to expand out the summation:
${\sum}_{n = 1}^{N} {n}^{4} = \left(\frac{1}{5}\right) {n}^{5} + \left(\frac{1}{2}\right) {n}^{4} + \left(\frac{1}{3}\right) {n}^{3} - \left(\frac{1}{30}\right) n$

${\lim}_{N \to \infty} {5}^{5} / {N}^{5} {\sum}_{n = 1}^{N} {n}^{4}$
$= {5}^{5} \cdot \frac{1}{5} + 0$
$= 625$

May 26, 2017

${\int}_{0}^{5} \setminus {x}^{4} \setminus \mathrm{dx} = 625$

#### Explanation:

By definition of an integral, then

${\int}_{a}^{b} \setminus f \left(x\right) \setminus \mathrm{dx}$

represents the area under the curve $y = f \left(x\right)$ between $x = a$ and $x = b$. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles.

That is

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

Here we have $f \left(x\right) = {x}^{4}$ and we partition the interval $\left[0 , 5\right]$ using:

$\Delta = \left\{0 , 0 + 1 \cdot \frac{5}{n} , 0 + 2 \cdot \frac{5}{n} , \ldots , 0 + n \cdot \frac{5}{n}\right\}$
$\setminus \setminus \setminus = \left\{0 , \frac{5}{n} , 2 \cdot \frac{5}{n} , \ldots , 5\right\}$

And so:

$I = {\int}_{0}^{5} \setminus {x}^{4} \setminus \mathrm{dx}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{5}{n} {\sum}_{i = 1}^{n} \setminus f \left(0 + i \cdot \frac{5}{n}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{5}{n} {\sum}_{i = 1}^{n} \setminus f \left(\frac{5 i}{n}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{5}{n} {\sum}_{i = 1}^{n} \setminus {\left(\frac{5 i}{n}\right)}^{4}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{5}{n} {\sum}_{i = 1}^{n} \setminus {\left(\frac{5}{n}\right)}^{4} {i}^{4}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{5}{n} \cdot \frac{625}{n} ^ 4 {\sum}_{i = 1}^{n} \setminus {i}^{4}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3125}{n} ^ 5 {\sum}_{i = 1}^{n} \setminus {i}^{4}$

Using the standard summation formula:

${\sum}_{r = 1}^{n} {r}^{4} = \frac{1}{30} \left(6 {n}^{5} + 15 {n}^{4} + 10 {n}^{3} - n\right)$

we have:

$I = {\lim}_{n \rightarrow \infty} \frac{3125}{n} ^ 5 \frac{1}{30} \left(6 {n}^{5} + 15 {n}^{4} + 10 {n}^{3} - n\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{625}{6} \left(6 + \frac{15}{n} + \frac{10}{n} ^ 2 - \frac{1}{n} ^ 4\right)$
$\setminus \setminus = \frac{625}{6} {\lim}_{n \rightarrow \infty} \left(6 + \frac{15}{n} + \frac{10}{n} ^ 2 - \frac{1}{n} ^ 4\right)$
$\setminus \setminus = \frac{625}{6} \left(6 + 0 + 0 - 0\right)$
$\setminus \setminus = 625$

Using Calculus

If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:

${\int}_{0}^{5} \setminus {x}^{4} \setminus \mathrm{dx} = {\left[{x}^{5} / 5\right]}_{0}^{5}$
$\text{ } = \frac{3125}{5} - 0$
$\text{ } = 625$

May 26, 2017

See below.

#### Explanation:

One of the possible realizations for that integral in terms of Riemann sum is

${\lim}_{n \to \infty} {\sum}_{k = 0}^{k = n} {\left(\frac{5 k}{n}\right)}^{4} \left(\frac{5}{n}\right)$

we have

${\lim}_{n \to \infty} {\sum}_{k = 0}^{k = n} {\left(\frac{5 k}{n}\right)}^{4} \left(\frac{5}{n}\right) = 625$