working out area under the curve?!!

• Nov 16th 2010, 02:40 PM
osmium
working out area under the curve?!!
the curve with equation y=x(6-x) crosses the x-axis at the origin O and at the point R. the rectangle OPQR is such that PQ is tangent to the curve at its maximum point.
1. show that the area of the region between the curve and the x-axis is two thirds of the area of the rectangle OPQR.

a straight line through the origin meets the curve at the point A with x-coordinate a.
2. find, in terms of a, the area of the shaded region between the curve and the line. hence find, correct to three significant figures, the value of a for which this area is half the area of the rectangle OPQR.
thank you!!
• Nov 16th 2010, 02:50 PM
Plato
Hello and welcome to MathHelpForum.
You should understand that this is not a homework service nor is it a tutorial service.
PLease either post some of your own work on this problem or explain what you do not understand about the question.
• Nov 16th 2010, 02:52 PM
osmium
i know the curve has the equation y=x(6-x). I integrated this to get (6x^2)/2- (x^3)/3. is this right? then i know that the graph crosses the x axis at the origin, and at x=6 i think. how do i then work out the area of the rectangle? because i need to prove that the area under the curve is 2/3 of the area of the rectangle. i worked out (when i subbed in 6 to the integration for the curve), that the area under the curve is 36. so surely the rectangle area is 54, but how?
• Nov 16th 2010, 03:18 PM
skeeter
Quote:

Originally Posted by osmium
i know the curve has the equation y=x(6-x). I integrated this to get (6x^2)/2- (x^3)/3. is this right? then i know that the graph crosses the x axis at the origin, and at x=6 i think. how do i then work out the area of the rectangle? because i need to prove that the area under the curve is 2/3 of the area of the rectangle. i worked out (when i subbed in 6 to the integration for the curve), that the area under the curve is 36. so surely the rectangle area is 54, but how?

parabola vertex is at x = 3 ... y = 3(6-3) = 9

rectangle area = 9(6) = 54