Let for some function y = f(x), 0x t f(t) dt = x2 f(x), x > 0 and f(2) = 3. Then f(6) is equal to
(1) 1
Given: 0x t f(t) dt = x2 f(x)
Differentiating both sides w.r.t. x (Leibniz rule):
x f(x) = 2x f(x) + x2 f'(x)
x2 f'(x) = -x f(x)
f'(x)f(x) = - 1x (x > 0)
|f(x)| = - x + C
f(x) = kx
f(2) = 3 k2 = 3 k = 6
f(x) = 6x
f(6) = 66 = 1
Answer: (1)
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