# What is the arclength of #f(x)=sqrt(4-x^2) # in the interval #[-2,2]#?

The arc length is

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To find the arc length of the function (f(x) = \sqrt{4 - x^2}) on the interval ([-2, 2]), we use the arc length formula:

[ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} , dx ]

In this case, (f(x) = \sqrt{4 - x^2}), so (f'(x) = -\frac{x}{\sqrt{4 - x^2}}).

Plugging these into the formula and evaluating the integral on ([-2, 2]), we get:

[ L = \int_{-2}^{2} \sqrt{1 + \left(-\frac{x}{\sqrt{4 - x^2}}\right)^2} , dx ]

[ L = \int_{-2}^{2} \sqrt{1 + \frac{x^2}{4 - x^2}} , dx ]

[ L = \int_{-2}^{2} \sqrt{\frac{4 - x^2 + x^2}{4 - x^2}} , dx ]

[ L = \int_{-2}^{2} \sqrt{\frac{4}{4 - x^2}} , dx ]

Now, perform a trigonometric substitution by letting (x = 2 \sin \theta). This leads to (dx = 2 \cos \theta , d\theta).

Substitute (x) and (dx) in terms of (\theta) into the integral:

[ L = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\frac{4}{4 - (2\sin\theta)^2}} \cdot 2\cos\theta , d\theta ]

[ L = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\frac{4}{4 - 4\sin^2\theta}} \cdot 2\cos\theta , d\theta ]

[ L = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\frac{4}{4(1 - \sin^2\theta)}} \cdot 2\cos\theta , d\theta ]

[ L = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2\cdot 2\cos\theta}{\sqrt{1 - \sin^2\theta}} , d\theta ]

[ L = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4, d\theta ]

[ L = 4\theta \bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} ]

[ L = 4\left(\frac{\pi}{2} + \frac{\pi}{2}\right) ]

[ L = 4\pi ]

So, the arc length of (f(x) = \sqrt{4 - x^2}) on the interval ([-2, 2]) is (4\pi).

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