Assume that 0< a < b <n infinity
Use geometric argument to show that integral a to b x^2 dx = (b^3-a^3/3)
I don't even understand what this is asking me - let alone a clue on how to attempt it.
Please help!
Nevermind - I figured this one out!!!!!!!!
Its a long one.
Sn = (a/n)f(0) = (a/n)f(a/n) = (a/n)f(2a/n)......(a/n)f((n-1)a/n)
= (a/n)0^2 + (a/n)(a^2/n^2) +.........(a/n)((n-1)^2a^2/n^2)
= a^3[1^2 = 2^2......(n-1)2]
Then you do this again but substitute a for b now to get the other side.
then
lim Sn = a^3/6(1-1/n)*1*(2 - 1/n) = a^3/6*1*1*2 = A^3/3
Now you do this for same thing but insert b.
Then you add a^3/3 + b^3/3 = (b^3 + a^3/3)