Calc ii

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\).

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

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  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"

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