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

##### 4 Answers

#### Answer:

#### Explanation:

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

#### Answer:

#### Explanation:

With limits you say... Ok.

Imaging we have some function

This means the height of each rectangle is equal to the value of the function at that point. Refer to

Therefore the area under

Where

We would also say that for smaller and smaller values of

Now that is some ugly stuff. Let's simplify this a bit and that advantage that the interval you want to find is

Another way of describing

Finally:

Oh, did I mention that

To do anything with it, we need to expand stuff. Starting with

And we need to expand out the summation:

#### Answer:

# 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

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

# 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:

See below.

#### Explanation:

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

we have