Question

a and b are positive integers and:
f(x) = x^n(a - bx)^n

1) What are the max and min values of function on interval [0, a/b]

2) Use the max value to give upper value for integral from 0 to a/b of f(x)

3) Explain why it is possible to find an n such that the integral in part 2 is less than 1

Answer 1

1) Since f_n(x) = (ax - bx^2)^n,
f '_n(x) = n(ax - bx^2)^(n-1) * (a - 2bx).

Setting this equal to 0 yields x = 0, a/b, a/(2b).

Testing these values (noting that x = 0, a/b are the endpoints of the interval as well):
f_n(0) = f_n(a/b) = 0 <----Minimum
f_n(a/(2b)) = (a^2/(4b))^n <----Maximum
---------------
2) ∫(x = 0 to a/b) f_n(x) dx < ∫(x = 0 to a/b) (a^2/(4b))^n dx = (a^2/(4b))^n * (a/b).

3) If we are assuming that a^2 < 4b (or even just a < b), then a^2/(4b) < 1
and hence (a^2/(4b))^n tends to 0 as n goes to infinity.

Hence, (a^2/(4b))^n * (a/b) < 1 for sufficiently large n.