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I need to evaluate $$\int_{-5}^5 \sqrt{25-x^2}~dx$$

How would I do it?

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If you are confused, you can do it without calculus. – André Nicolas Jul 12 '13 at 18:15
Have you heard of trigonometric substitution? – Jonathan Dewein Jul 12 '13 at 18:17
Some could say that being into integration is not a good place to be "so confused" about calculus any more. This can happen, and it usually happens, at the beginning: functions, limits, continuity, derivatives...but upon reaching Riemann integrals either one already gave up or got into business. – DonAntonio Jul 12 '13 at 23:26
Its overkill, but you can try an Euler's Substitution of the second type. – Gamma Function Jul 13 '13 at 1:14
up vote 11 down vote accepted

If you are at the earliest stages of integration, you are probably supposed to do this problem without using any integration techniques.

The curve $y=\sqrt{25-x^2}$ is the top half of a circle with centre the origin and radius $5$.

Our integral $\int_{-5}^5 \sqrt{25-x^2}\,dx$ is the area under the curve $y=\sqrt{25-x^2}$ and above the $x$ axis, from $x=-5$ to $x=5$. That's a half-circle. Thus by a familiar formula the area is $25\pi/2$.

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Thank you! @Andre Nicolas – a12b23c34 Jul 12 '13 at 20:26
You are welcome. Had to guess what you were looking for and at what stage you were in integration. If you have already covered trigonometric substitutions, and the integration of $\cos^2\theta \,d\theta$, then James Maslek's answer is perhaps more appropriate. The usual way to handle $\cos^2\theta$ is to use the identity $\cos 2\theta=2\cos^2\theta -1$. So you end up integrating $\frac{1+\cos 2\theta}{2}$. There are other ways to integrate $\sqrt{25-x^2}$, for example integration by parts. – André Nicolas Jul 12 '13 at 20:38

$$\int_{-5}^5 \sqrt {25 - x^2} \,dx = \text{area of semi-circle of radius 5}$$

That is, $y = \sqrt{25 - x^2},\;$ defines the top half of the circle $x^2 + y^2 = 5^2 = 25,\;$ which is centered at the origin, and has a radius of $5$. See the "real" portion of the graph below, outlined in blue:

enter image description here

If you are taking calculus, then we need for you to get comfortable with computing area using integrals! Short cuts won't help in the long run. Besides, it's fun, once you get the hang of it.

So let's do Calculus!: you can integrate the integral fairly easily by using trigonometric substitution.

In this case, let's put $\,\bf x = 5\sin \theta$. Then $\bf\,dx = 5 \cos \theta\,d\theta.\,$ and $\theta = \sin^{-1}\left(\dfrac x5\right).$ Our bound of integration then become $(\theta = 3\pi/2 \;\text{to} \;\theta =\pi/2)$. But we'll simplify matters by doubling the area of the circle as theta sweeps from $0$ to $\pi/2$.

So, $$\begin{align}\int \sqrt{25 - x^2}\, dx & = \int \left(\sqrt{25 - 25\sin^2 \theta}\right)(5\cos \theta)\,d\theta \\ \\ & = 5\int \left(\sqrt{25(1 - sin^2 \theta}\right)\cos \theta\,d\theta\\ \\ & = 5\int 5\sqrt{\cos^2x} \cos \theta\,d\theta\\ \\ &= 25 \int \cos^2 \theta\,d\theta \\ \\ & = 25\cdot 2\int_0^{\pi/2}\left( \dfrac 12 + \dfrac {\cos 2\theta}2\right)\,d\theta\end{align}$$

In the last step we simply use the identity: $$\cos^2 \theta = \dfrac{1 + \cos 2 \theta}{2}$$. Then it's just a matter of using the power rule, and you're almost there.

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Thanks, Amzoti! – amWhy Jul 13 '13 at 1:07

Easy way is what @AndreNicolas said. Calculus approach is trig substitution (here is a link)

So we let

Let x = $5\sin(\theta)$ , $dx = 5\cos(\theta)d\theta$


So your integral becomes $\int^{\sin^{-1}(1)}_{\sin^{-1}(-1)}25\cos^2(\theta)d\theta$. When you do this out you get 25$\pi$/2. Coincidence?

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The bounds change because you have to account for this when you change x to $\theta$ – yankeefan11 Jul 12 '13 at 18:52
My bad. The lower bound needed to be $\sin^{-1}(-1)$ – yankeefan11 Jul 12 '13 at 19:35
I have a question. How would you get the 25pi/2 once you have your integral of ∫sin−1(1)sin−1(−1)25cos2(θ)dθ? What do the two values come out to be once you plug them in, so what minus what equals 25pi/2? – a12b23c34 Jul 12 '13 at 20:23
And thank you!@James Maslek – a12b23c34 Jul 12 '13 at 20:25

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