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Feb28-12, 09:19 AM
Find the area of the loop of the curve y^2=x^3(1-x)^2 using integral calculus.
The following needs extra parentheses to be correct.
This should be y = ±√(x
) (1-x) or y = ±x
To sketch the curve, I assigned values for x and then solved the corresponding values of y.
x= -1, y= -2
x= -0.5, y= -0.53
x=0, y= 0
x= 0.5, y= 0.177
how can i find the area of this? >.<
Hello stardust006. Welcome to PF !
The ± is important. It gives a clue as to why the graph has a loop.
There are two x-intercepts. Can you find them ?