00:09
By the remandum term.
00:11
Remenson be a fx here and we just divide fx in block of rectangles we have so that we take here uh we have fx we take and then we take delta x deltac here we have and the value is the fx here this we take and we'll summation accordingly to summation uh we do this so if the remand term how do we evaluate this so take an fx i from i equal to 0 to n negative 1 that means when the area under this curve we're adding here the block of rectangles from i equal to 0 to n negative 1 and if we apply here limit and 10 to infinity we have limit and 10 infinity that means we just make it as closer as possible in block of a rectangle so that it just become like a point and this here we have this become like a point and this vertical distance another point in the vertical distance another point in the vertical distance so we have n tends to infinity we get the complete area under the curve so formation i equal to 0 to n negative 1 f x i delta if there's two this limit n density we get integral f x d x that's what we go the integral apex d x and that we can take four let's say you start from here a to we can take that integral from a to b in this way we can divide it we just work on the given limit now so we can take it as summation k equals 0 n negative 1 logo 1 negative k over n if we just work on this we are written now here we have log 1 plus logg 1 over n and all this we have is going to be logo one negative n negative 1 over n this way we can just put the summation and then it is multiplied with one over and the given quotient malady 1 over n so here we have multiply with 1 over n 1 over n 1 over and here we have 1 over let's graph it and try to understand it using the to see now, we've taken from 0 to 1, low x to be taken.
02:46
Now to find the area under this third, what we can do here to find the area under this curve, the divide in the block of rectangles here we have.
02:54
So one rectangle is log 1 negative 2 over n times 1 over n.
02:58
This is log 1 times 1 over n.
03:00
This is log 1 negative 3 over n.
03:05
I'm sorry, this is log 1 negative 2 over n times 1 over and.
03:08
Or even if we can take here, this will be low 1 times 1 over n.
03:10
1 negative 1 over 1 over 1.
03:13
The first one is log 1 times 1 over n.
03:16
Second one is log 1 negative 1 over n times 1 over n.
03:19
3rd 1 is this one here log 1 negative 2 over n times 1 over and so on till log 1 negative and negative 1 over n.
03:30
So if we just put the limit here limit and 10 infinity, even the whole if it's put limit and 10 infinity...