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In my textbook appears that $\displaystyle\int_{-1}^{1}\sqrt{1-x^2}dx=\frac{\pi}{2}$

But where does this equation come from?

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    I did a grammatical edit. In English one may say "This comes from X" or "Where does this come from", or "This came from X" or "Where did this come from", but never "Where does this comes from" or "Where did this came from".2012-05-10

4 Answers 4

10

The equation $y = +\sqrt{1 - x^2}$ for $-1\leq x\leq1$ describes the top half of a circle of radius $1$. The area between this curve and the $x$-axis is therefore $\pi/2$. On the other hand, you can compute the area under this curve by doing the integral $\int_{-1}^1 \sqrt{1 - x^2}\,dx.$

7

Those "area" answers are probably the best ones. On the other hand, if $\pi$ is defined in some way other than area, we can pursue the standard trigonometric substitution: $x = \sin \theta$, $-\pi/2 \le \theta \le \pi/2$ to get: $\begin{align} \int_{-1}^1\sqrt{1-x^2}\,dx &= \int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta = \int_{-\pi/2}^{\pi/2}\frac{1+\cos(2\theta)}{2}\,d\theta \\ &=\frac{1}{2}\int_{-\pi/2}^{\pi/2}d\theta + \frac{1}{2}\int_{-\pi/2}^{\pi/2} \cos(2\theta)\,d\theta \\ &= \frac{\pi}{2} + 0 = \frac{\pi}{2} . \end{align}$

5

Not the "cleverest" method, like the above, - but works!

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

To compute that integral, one may substitute $x=\sin{t}$, and get:

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1-\sin^2{t}} \ d(\sin{t})=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos{t}\cos{t} \ dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2{t} \ dt$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2{t} \ dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1+\cos{2t}}{2} \ dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dt}{2}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos{2t}}{2}=$

$= \frac{\pi}{2}+\frac{1}{2}\sin{t}\cos{t}{\huge{|}}_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{\pi}{2}+0=\frac{\pi}{2}$

4

$f(x)=\sqrt{1-x^2}$ is the graph of the upper half of the circle $x^2+y^2=1$. To see this, just square both side.

Thus,

$\int_{-1}^1 \sqrt{1-x^2}dx$ is the integral which represents the area under half of circle of radius 1. That is the are of half disk, thus $\frac{1}{2} \pi 1^2$