1. ## antiderivative of ...

I'm trying to show the calculation of the area under the graph of semicircle f(x) = - (4-x^2)^1/2 from [-2,2] by integration.

I understand the calculation can be done by taking antiderivative F(2) - F (-2), but I can't remember how to find F for this type.

Thanks!

2. Do you mean how to setup the integral?

$\displaystyle 2 \int_0^2 \sqrt{4-x^2}~dx$

or how to actually integrate it?

3. Sorry, yes how to integrate it. I remember how to integrate simple polynomials, but not square roots with exponents.

4. $\displaystyle 2 \int_0^2 \sqrt{4-x^2}~dx$

=$\displaystyle 2 \int_0^2 2\sqrt{1-\bigg(\frac{x}{2}\bigg)^2}$

=$\displaystyle 4 \int_0^2 \sqrt{1-\bigg(\frac{x}{2}\bigg)^2}$

Ok now we need some substitutions and a trig identity.

$\displaystyle 1- sin^2(u) = cos^2(u)$

$\displaystyle sin(u) = \frac{x}{2}$

$\displaystyle dx = 2cos(u)$

Also have to change upper and lower bounds

new lower bound

$\displaystyle \frac{0}{2} = sin(u)$

new upper bound

$\displaystyle \frac{2}{2} = sin(u)$

so now with the substitutions

$\displaystyle 8 \int_0^{\frac{\pi}{2}} cos^2(u)~du$ = $\displaystyle 8 \int_0^{\frac{\pi}{2}} \bigg(\frac{1 + cos(2u)}{2}\bigg)~du$

= $\displaystyle 4u + 2sin(2u)\bigg|_0^{\frac{\pi}{2}}$ = $\displaystyle \frac{\pi}{2}$

5. Sub $\displaystyle x = 2\sin{\theta}$

6. Originally Posted by 11rdc11
$\displaystyle 2 \int_0^2 \sqrt{4-x^2}~dx$

=$\displaystyle 2 \int_0^2 2\sqrt{1-\bigg(\frac{x}{2}\bigg)^2}$

=$\displaystyle 4 \int_0^2 \sqrt{1-\bigg(\frac{x}{2}\bigg)^2}$

Understand so far?
Nope, sorry. I'm so rusty that I would really be grateful for some sentences. I can integrate polynomials no problem, but that's about it.

7. What part do you need me to explain? Feel free to ask questions.

8. Originally Posted by 2clients
Nope, sorry. I'm so rusty that I would really be grateful for some sentences. I can integrate polynomials no problem, but that's about it.
If you sub $\displaystyle x = 2\sin{\theta}, dx = 2\cos{\theta} d\theta$, then;

$\displaystyle 4 \int 2\cos{\theta}\sqrt{4-4\sin^2{\theta}}~d\theta$

$\displaystyle 16 \int \cos^2{\theta}~d\theta$

And use the half angle identity $\displaystyle \cos^2{\theta} = \frac{1}{2}(1+\cos{2\theta})$. Problem solved.

9. Oops didn't think about that, guess that what happens when your sleepy