Let f be twice differentiable with (f(0)=4), (f(1)=8), and (f'(1)=6). Evaluate the integral from 0 to 1 of xf''(x)dx\).

Question

I have no clue how to approach this since we don't know what the function is? please help, I've finished every other question and the assignemnt is due at 9

Answer 1

∫[0,1] xf"(x) dx

Integrate by parts.
Let u = x
Let dv = f"(x)
then
du = dx
v = f'(x)

∫u dv = uv - ∫v du
= xf'(x) [0,1] - ∫[0,1]f'(x) dx
= xf'(x) [0,1] - f(x) [0,1]
= (xf'(x)-f(x))[0,1]
= (1*f'(1)-f(1))-(0*f'(0)-f(0))
= (6-8)-(0-4)
= 2

check:
d/dx xf'-f = f' + xf" - f' = xf"

Answer 2

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