How would you solve the integral sqrt(a^2-x^2) from -a to a in the simplest easiest way ? =)

Question

lgbasallote · Answer

if i recall right....there was this law that says $$\int \limits_{-a}^a (\text{even fuction} )dx = 0$$
or was it for the odd function...

i'll try and check

lgbasallote · Answer

hmm lol seems it's for the odd function

anonymous · Answer

lol !! but at least you reminded me this rule as well ;)

anonymous · Answer

the answer should be pi*a^2/2 but i can't get it right =\

lgbasallote · Answer

ahh i got it

lgbasallote · Answer

$$\Large \int\limits_{-a}^a (a^2 - x^2)dx \implies 2\int\limits_{-a}^a x^2 dx$$
does that help?

anonymous · Answer

you mean twice the integral from 0 to $a$

anonymous · Answer

oh it is the square root!!!

anonymous · Answer

$$\int_{-a}^a\sqrt{a^2-x^2}dx$$ like that i bet

anonymous · Answer

do not integration at all

anonymous · Answer

$y=\sqrt{a^2-x^2}$ represents a circle, center at the origin, radius of $a$
geometry tells you that the area of the circle is $\pi a^2$
since you are taking half of that circle, area is 
$$\frac{\pi a^2}{2}$$

anonymous · Answer

i meant of course the upper half of the circle |dw:1346161966158:dw|

anonymous · Answer

Great !!!!! thank you !!!!!!! =) Nice thinking !!!