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**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?