00:01
Let's talk about question number 62.
00:02
So we've got to find the function whose derivative is this and which passes in the function passes through 2 .7.
00:08
So let's integrate this both sides.
00:11
So now we have fx because integration of f -dash -x is just a fixed.
00:16
So we have fx is equal to integration of negative 2x square root of 8 minus x squared dx.
00:23
So let's substitute.
00:26
I'm going to use a u substitution over here.
00:28
So 8 minus x squared is substituted as u.
00:31
So if we differentiate both sides, differentiation of 8 is just 0 and of negative x squared is negative 2x dx, and this is equal to du.
00:40
So i think we are good to apply the integration and replace everything in terms of u.
00:47
So the integration becomes 8 minus x square is replaced by u.
00:51
So we have root of u, and negative 2x d x is just d u.
00:55
So we have integration of u raised to one over two d u, which is nothing but u raised to 3 over 2 over 2 plus c...