# Finding the area between $x \sqrt{4x-x^2}$ and $\sqrt{4x-x^2}$

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?

• just wonder will $y_1$, $y_2$ be not differentiable but integrable at $x=0/4$. – Mythomorphic Jun 23 '15 at 7:19

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 $2$.