limit of intergral question. [0,1] intergral x/sqroot[4-x^2]

[kpx001]

ok so basiclly what i am asking is what is a and b.
i did the work using trig sub. x= 2sin@ du = 2cos@

2sin@/sqroot[4-4sin^2@] => 2*intergral sin@ . my only question is what are the limits? the answer is 0 to pi/6 , but i dont know how they got that.

 

[galactus]

Just by plugging their old limits into the sub and solving for theta.

\(\displaystyle x=2sin{\theta}\Rightarrow \;\ {\theta}=sin^{-1}(\frac{x}{2})\)

\(\displaystyle sin^{-1}(\frac{1}{2}), \;\ sin^{-1}(0)\)

This gives new limits of \(\displaystyle 0, \;\ \frac{\pi}{6}\)

 

[fasteddie65]

Actually, you don't need to make a trig substitution at all.

Let u = 4 - x^2
du = -2x dx
(-1/2) du = x dx

The integral then becomes (-1/2) int [du/sqrt u] = (-1/2) int [u^(-1/2) du]

I'll bet you can finish it from here. (Don't forget to change the limits to u-limits instead of x-limits.)