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We are trying to evaluate

$$ \int\limits_0^1 ( \sqrt{2-x^2} - \sqrt{2x-x^2} ~) ~ dx $$

without any substitution (well, this is how this problem is supposed to be solved)

Idea:

We notice that if $y=f(x)$ is the integrand, then $f(1) = \sqrt{2}$ and $f(1)=0$ as is evident. So, my idea would be to evaluate

$$ \int\limits_0^{\sqrt{2}} f^{-1}(y) ~ dy $$

But, this would make it harder since we would need to solve $y = \sqrt{2-x^2} - \sqrt{2x-x^2}$ for $x$... Any ideas how would we tackle this problem?

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ILoveMath
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1 Answers1

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One possible approach to solve this problem consists in interpreting integrals as areas. Indeed, consider the functions \begin{align*} \begin{cases} f(x) = \sqrt{2 - x^{2}}\\ g(x) = \sqrt{2x - x^{2}} = \sqrt{1 - (x-1)^{2}} \end{cases} \end{align*}

The graph of the function $f$ is the upper semicircle centered at the origin with radius $r_{1} = \sqrt{2}$. Thus the area corresponding to the first integral is given by \begin{align*} \int_{0}^{1}f(x)\mathrm{d}x = \frac{\pi r_{1}^{2}}{4} - \left(\frac{\pi r_{1}^{2}}{8} - \frac{1}{2}\right) = \frac{\pi r_{1}^{2}}{8} + \frac{1}{2} = \frac{\pi}{4} + \frac{1}{2} \end{align*}

On the other hand, the graph of the function $g$ is the upper semicircle centered at $(1,0)$ whose radius equals $r_{2} = 1$. Therefore we have \begin{align*} \int_{0}^{1}g(x)\mathrm{d}x = \frac{\pi r_{2}^{2}}{4} = \frac{\pi}{4} \end{align*}

Finally, we have \begin{align*} \int_{0}^{1}[f(x) - g(x)]\mathrm{d}x = \frac{1}{2} \end{align*}

user0102
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