Tuesday

September 23, 2014

September 23, 2014

Posted by **Raven** on Sunday, November 28, 2010 at 12:43pm.

- calculus -
**MathMate**, Sunday, November 28, 2010 at 1:07pmThe 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

**Answer this Question**

**Related Questions**

calculus - 1. Find the area of the region bounded by f(x)=x^2 +6x+9 and g(x)=5(x...

Calculus - Find the area of the region bounded by the graphs of the given ...

Calculus AP - Let R be the region bounded by the graphs of y=cos((pi x)/2) and y...

calculus - Sketch the region bounded by the graphs of the algebraic functions & ...

calculus - FInd the area of the region bounded by the graphs of y= 3x^2+1 , y=0...

calculus - Find the area of the region bounded by the graphs f(x)=sqrt(3x)+1, g(...

calculus - find the volume of the solid bounded above by the surface z=f(x,y) ...

Calculus AB/AP - Use the rectangles with the given dimensions to approximate the...

calculus - Consider the graphs of y = 3x + c and y^2 = 6x, where c is a real ...

calculus - Find the area of the region bounded by the graphs of the algebraic ...