00:01
Hi, let's start the solution.
00:02
In this question, we have given for 1962, the lorentz curve of x is equal to 2 by 5 x plus 3 by 5 x square.
00:28
So, then gini index for 1962 is given by 2 integration 0 to 1 x minus f of x dx.
00:45
So, we get that is 2 integration 0 to 1 x minus f of x is 2 by 5 x minus 3 by 5 x square dx.
00:56
That is equal to 2 integration 0 to 1 5 x minus 2 x divided by 5 minus 3 by 5 x square dx.
01:08
So, we get that is equal to 2 integration 0 to 1 3 by 5 x minus 3 by 5 x square dx.
01:22
Now, by integrating this here, we get 2 3 by 5 x square by 2 minus 3 by 5 x cube by 3 limit 0 to 1.
01:36
Now, by taking 2 into 3 by 5 common, we get x square by 2 minus x cube by 3.
01:46
That is we have 6 by 5 limit 0 to 1.
01:53
Now, apply the limit.
01:53
So, we get 1 by 2 minus 1 by 3 that is 6 by 5 by taking lcm, we get 6 in numerator, we get 3 minus 2...