integrate sqrt(16-y^2)/2 dy

Question

shamil98 · Answer

$$\large \sqrt{16-y^2} = (16-y^2)^\frac{ 1 }{ 2 }$$

shamil98 · Answer

There isn't a rule for integration like the quotient rule is there?..

anonymous · Answer

integration by parts or integration by subtitution

shamil98 · Answer

yeah, i'm looking at that right now.

shamil98 · Answer

First choose u and v
u = (16-y^2)^1/2
v = 1/2

shamil98 · Answer

differentiate u and integrate v.

shamil98 · Answer

ya with me so far?

shamil98 · Answer

we're using the formula
$$\large \int\limits_{}^{} udv = uv - \int\limits_{}^{} v~du$$

anonymous · Answer

erm, wait me read up for a while >< coz preparing test for other subject tmr, n doing my engineering math assignment simultaneously

shamil98 · Answer

use the chain rule for u and integrate v 
(16-y^2)^1/2 = u
u' = -2y/(16-y^2)^1/2
v = 1/2
integrating v = x/2 + c

shamil98 · Answer

I'm just following an example lol , not sure if i'm doing this correctly >.<

anonymous · Answer

ok, nvm, we can study together coz i left my integration by parts and integration by substitution abt 1 year ald

anonymous · Answer

i get the fundamental of integration by part ald