find the area the region bounded by the graphs of the given equation y=81x,y=x^5,x=0,x=3

The area is between the function y=x^5 and y=81x, between the limits x=0 and x=3.

The functions are all in the first quadrant (no values negative), and the two curves y=81x and y=x^5 intersect at x=3, the upper integration limit, which is a "lucky coincidence".

The area function a(x), is therefore
a(x)=81x-x^5
You would integrate a(x) from 0 to 3 to find the area, thus
Area = ∫a(x)dx [x=0,3]
=∫(81x-x^5)dx [x=0,3]

Here's an image to give you an idea of the area:
http://img11.imageshack.us/img11/6060/1290966183.png

To find the area of the region bounded by the given graphs, we need to determine the points of intersection first. Once we have the points of intersection, we can integrate the two functions and subtract one from the other to find the area.

Let's find the points of intersection:
1. Set the two given functions equal to each other to find where they intersect:
y = 81x and y = x^5
Setting 81x equal to x^5: 81x = x^5

2. Simplify the equation:
Subtracting 81x from both sides: x^5 - 81x = 0

3. Factor the equation:
Factoring out an x: x(x^4 - 81) = 0
Applying difference of squares to the second part: x(x^2 + 9)(x^2 - 9) = 0
Simplifying further: x(x^2 + 9)(x + 3)(x - 3) = 0

So, the potential points of intersection are x = 0, x = ±3.

Now, let's integrate the two functions and evaluate the definite integral between these intersection points to find the area:

Area = ∫[lower bound, upper bound] (81x - x^5) dx

1. For the interval between x = 0 to x = 3:
Area = ∫[0, 3] (81x - x^5) dx

Integrating the function:
Area = [40.5x^2 - (x^6)/6] evaluated from 0 to 3
Area = [(40.5 * 3^2) - (3^6)/6] - [(40.5 * 0^2) - (0^6)/6]
Area = [40.5 * 9 - 729/6] - [0]
Area = 364.5 - 121.5
Area = 243 square units

Therefore, the area of the region bounded by the graphs of the given equations is 243 square units.