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

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:

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See below.

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

we have

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

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

By definition of an integral, then

That is

And so:

Using the standard summation formula:

we have:

Using Calculus

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

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To find the area between the curve ( y = x^4 ) and the x-axis from ( x = 0 ) to ( x = 5 ), you can integrate the absolute value of the function over the given interval. The area is given by the integral:

[ \text{Area} = \int_{0}^{5} |x^4| , dx ]

You need to break this integral into two parts because ( x^4 ) changes sign at ( x = 0 ). The integral from ( x = 0 ) to the point where ( x^4 = 0 ) (which is ( x = 0 )) gives the area under the curve from ( x = 0 ) to the x-axis, and the integral from that point to ( x = 5 ) gives the area above the x-axis.

So, you can rewrite the integral as:

[ \text{Area} = \int_{0}^{5} x^4 , dx - \int_{0}^{0} x^4 , dx ]

Evaluate each integral separately:

[ \int_{0}^{5} x^4 , dx = \frac{1}{5}x^5 \bigg|_{0}^{5} = \frac{1}{5}(5)^5 - \frac{1}{5}(0)^5 = \frac{3125}{5} = 625 ]

[ \int_{0}^{0} x^4 , dx = 0 ]

So, the total area is ( 625 ) square units.

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