Proof that $\int_{-1}^{1}\sqrt{1-x^2}dx=\frac{\pi}{2}$?

Question

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

But where does this equation come from?

Answer 1

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

Answer 2

$f(x)=\sqrt{1-x^2}$ 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 $\frac{1}{2} \pi 1^2$

Answer 3

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: $$\begin{align} \int_{-1}^1\sqrt{1-x^2}\,dx &= \int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta = \int_{-\pi/2}^{\pi/2}\frac{1+\cos(2\theta)}{2}\,d\theta \\ &=\frac{1}{2}\int_{-\pi/2}^{\pi/2}d\theta + \frac{1}{2}\int_{-\pi/2}^{\pi/2} \cos(2\theta)\,d\theta \\ &= \frac{\pi}{2} + 0 = \frac{\pi}{2} . \end{align}$$

Answer 4

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:

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

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

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