So, the task is to calculate the area of a shape in xOy plane bounded by functions: $y = x\sqrt{4x-x^2}$ and $y = \sqrt{4x-x^2}$ Could you please explain how I can solve this? How can I find the intersection points of these functions and how do I know which function has a bigger y value for the range of x values.
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$\begingroup$ First step: Graph both functions in one xy plane with a graphing utility $\endgroup$– imranfatJun 27, 2015 at 21:21
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2 Answers
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The domain of both functions is given by $$4x-x^2=x(4-x)\ge0,$$i.e.$$0\le x\le4.$$
Then the intersections are such that $$x\sqrt{x(4-x)}=\sqrt{x(4-x)}.$$
This occurs at $x=1$ and the endpoints of the domain, $x=0,x=4$.
Obviously, $LHS\ge RHS$ when $x\ge 1$.
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Hint:
Blue $\rightarrow y=\sqrt{4x-x^2}$
Purple $\rightarrow y=x\sqrt{4x-x^2}$
answered Jun 27, 2015 at 21:33