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So I've been doing real analysis for a last couple of days, and stumbled upon this task. The task is to find the area enclosed by $$y_1=x\sqrt{4x-x^2} $$ and $$y_2= \sqrt{4x-x^2} $$ This is one of those problems I see and think to myself "My God do I have to integrate that function..". I started of by analyzing both functions and made some conclusions.

  • Both functions have a domain $D=[0,4]$

  • None of the functions express asymptotic behaviour

    • For $0 \le x \le 1$ it holds that $y_2>y_1$

    • For $1 < x \le 4$ it holds that $y_1>y_2$

    • Functions are not differentiable at 0 and 4

To sum up, the area is given by

$$ A= \int_{0}^{1} (y_2-y_1) dx + \int_{1}^{4} (y_1-y_2)dx $$

So here comes my problem. The integration of these functions is doable, but the possibility of a mistake is great. Also, when solved, there is a 0 in the denominator of one of the fractions ( at least the form I got it in) because the function iz not differentiable in those points. So my questions are : Is it possible to deal with this integration on a more elegant way and what to do when the function approaches 0 or 4. I understand I could take right limit at 0 and left limit at 4, but is that correct?

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  • $\begingroup$ just wonder will $y_1$, $y_2$ be not differentiable but integrable at $x=0/4$. $\endgroup$ – Mythomorphic Jun 23 '15 at 7:19
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It’s easier if you shift the horizontal axis.

If you make the substitution $x=u+2$, you’re looking at the functions $f(u)=\sqrt{4-u^2}$ and $g(u)=(u+2)\sqrt{4-u^2}$ for $-2\le u\le 2$, which are equal at $u=-2,-1$, and $2$. The differences are

$$g(x)-f(x)=(u+1)\sqrt{4-u^2}=u(4-u^2)^{1/2}+\sqrt{4-u^2}$$

and its negative. Integrating the first term on the right is straightforward, and the integral of the second from $-2$ to $2$ is on geometric considerations the area of a semicircle of radius $2$.

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