Question

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

Answer 1

`625`

Explanation:

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

`int_0^5[x^4]dx=x^5/5=5^4=625`

Answer 2

`625`

Explanation:

With limits you say... Ok.

Imaging we have some function `f(x)` and we want to find the area underneath it in the interval `[a, b]`. 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(x_0)`, and the area of it is equal to `f(x_0)Delta x` where `Delta x` is just the constant width of each rectangle.

Therefore the area under `[a, b]` is in the diagram above is:
`sum_(n=0)^N R_n=sum_(n=0)^N f(a+nDeltax) Deltax=Deltax sum_(n=0)^N f(a+nDeltax)`
Where `N=(|a-b|)/(Deltax)`

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!

`Area=lim_(Delta x -> 0) sum_(n=0)^N f(a+nDeltax) Deltax`

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

`Area=lim_(Delta x -> 0) sum_(n=0)^N f(nDeltax) Deltax`

Another way of describing `Delta x` is `(|a-b|)/N` (see above equations). So, in this case, it's `(|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 -> oo)`
Finally:
`Area=lim_(N -> oo) sum_(n=1)^N f(5/Nn) 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(x)` and then factoring out any things that can be factored:

`lim_(N -> oo) sum_(n=1)^N f(5/Nn) 5/N=lim_(N -> oo) 5/N sum_(n=1)^N (5/N)^4n^4`
`=lim_(N -> oo) 5^5/N^5 sum_(n=1)^N n^4`

And we need to expand out the summation:
`sum_(n=1)^N n^4=(1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n`

`lim_(N -> oo) 5^5/N^5 sum_(n=1)^N n^4`
`=5^5 * 1/5 + 0`
`=625`

Answer 3

`int_0^5 \ x^4 \ dx = 625`

Explanation:

By definition of an integral, then

`int_a^b \ f(x) \ dx`

represents the area under the curve `y=f(x)` 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 \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)`

Here we have `f(x)=x^4` and we partition the interval `[0,5]` using:

`Delta = {0, 0+1*5/n, 0+2*5/n, ..., 0+n*5/n }`
`\ \ \ = {0, 5/n, 2*5/n, ..., 5 }`

And so:

`I = int_0^5 \ x^4 \ dx`
`\ \ = lim_(n rarr oo) 5/n sum_(i=1)^n \ f(0+i*5/n)`
`\ \ = lim_(n rarr oo) 5/n sum_(i=1)^n \ f((5i)/n)`
`\ \ = lim_(n rarr oo) 5/n sum_(i=1)^n \ ((5i)/n)^4`
`\ \ = lim_(n rarr oo) 5/n sum_(i=1)^n \ (5/n)^4i^4`
`\ \ = lim_(n rarr oo) 5/n * 625/n^4sum_(i=1)^n \ i^4`
`\ \ = lim_(n rarr oo) 3125/n^5sum_(i=1)^n \ i^4`

Using the standard summation formula:

`sum_(r=1)^n r^4 = 1/30(6n^5+15n^4+10n^3-n)`

we have:

`I = lim_(n rarr oo) 3125/n^5 1/30 (6n^5+15n^4+10n^3-n)`
`\ \ = lim_(n rarr oo) 625/6 (6+15/n+10/n^2-1/n^4)`
`\ \ = 625/6 lim_(n rarr oo) (6+15/n+10/n^2-1/n^4)`
`\ \ = 625/6 (6+0+0-0)`
`\ \ = 625`

Using Calculus

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

`int_0^5 \ x^4 \ dx = [ x^5/5 ]_0^5`
`" " = 3125/5-0`
`" " = 625`

Answer 4

See below.

Explanation:

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

`lim_(n->oo)sum_(k=0)^(k=n) ((5k)/n)^4(5/n)`

we have

`lim_(n->oo)sum_(k=0)^(k=n) ((5k)/n)^4(5/n)=625`