integration Q (1 Viewer)

[jkwii]

just started integration...
prove that
2
∫ √(x(4-x)) dx = pi
0

 

[KennethLiu]

well... first complete the square of x(4-x) so it becomes 4-(x-2)2

then substitute x-2=2sin a,
x = 2sin(a) +2,
dx = 2cos(a) da

bounds:
x: 0~2,
a: -pi/2~0

anyway then sub it back in, the function to be integrated should be

∫ √[4-4(sin a)squared] da
∫ 2 (cos a)squared da

hope u get that.. lol cant type squared here... and dont 4get to change the bounds, i am just too lazy to type that much...XD

anyway then use the double angle formula of cos2a, so the above becomes this...

2 ∫ (1+cos 2a) da

and integrate it... 2 [a + (1/2) sin 2a]

sub in the bounds and u should be able 2 get pi

 

[doink]

Its a semicircle.