# Integral $\int_{0}^{1}(\sqrt{2-x^2}-\sqrt{2x-x^2})dx$

I am working on the following integral problem:

$$\int_{0}^{1}(\sqrt{2-x^2}-\sqrt{2x-x^2})dx$$

There is a hint as well, which suggests interpreting the definite integral as the area bounded by appropriate curves. I have graphed both the curves with Wolfram Alpha (link), and I understand the area I am trying to find. I am stuck on solving the integral. The tools I have been provided with so far are fairly basic such as substitution and integration by parts.

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For first integral, let $x=\sqrt{2}\sin\theta$. For second, note that $2x-x^2=1-(1-x)^2$ and let $1-x=\sin\phi$. – André Nicolas Jan 12 '12 at 5:17

## 1 Answer

André has given you a calculus solution; here’s one that uses no calculus.

The graph of $y=\sqrt{2-x^2}$ is the upper half of a circle of radius $\sqrt2$ centred at the origin; that of $y=\sqrt{2x-x^2}=\sqrt{1-(x-1)^2}$ is the upper half of a circle of radius $1$ centred at $(1,0)$. Thus, $$\int_0^1 \sqrt{2x-x^2} dx=\frac{\pi}4\;,$$ a quarter of the area of a circle of radius $1$. We can also get $$\int_0^1 \sqrt{2-x^2} dx$$ without calculus, but it requires a little more cleverness. If you sketch the quarter-disk in the first quadrant bounded by $y=\sqrt{2-x^2}$, you’ll see that it’s the union of the unit square $$S=\{(x,y):0\le x,y\le 1\}\;,$$ the region $T$ bounded by the $y$-axis, the line $y=1$, and the curve $y=\sqrt{2-x^2}$, and the region $R$ bounded by the $x$-axis, the line $x=1$, and the curve $y=\sqrt{2-x^2}$. Regions $T$ and $R$ are clearly congruent, so they have the same area. The area of the whole quarter-disk is $\pi/2$, so $$\frac{\pi}2=1+2\operatorname{area}(T)\;,$$ and $$\operatorname{area}(T)=\frac12\left(\frac{\pi}2-1\right)=\frac{\pi-2}4\;.$$ Finally, $$\int_0^1 \sqrt{2-x^2} dx=\operatorname{area}(S)+\operatorname{area}(T)=1+\frac{\pi-2}4=\frac{\pi+2}4\;,$$ and $$\int_{0}^{1}\left(\sqrt{2-x^2}-\sqrt{2x-x^2}\right)dx=\frac{\pi+2}4-\frac{\pi}4=\frac12\;.$$

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@J.Borges: If you are interested in the history of mathematics, you might like to know that this is essentially the famous quadrature of a lune by Hippocrates. – André Nicolas Jan 12 '12 at 5:52
Looking at the figure described in the first three lines one immediately sees that the area in question is $\bigl(({1\over 8}2\pi+{1\over 2}\bigr) - {\pi\over 4}={1\over 2}$. – Christian Blatter Jan 12 '12 at 14:14
Thank you very much for your help Brian, and thank you very much André for the link (I'll have to wait until Wikipedia is back online, however). – J.Borges Jan 18 '12 at 19:52