Suppose that f(1)=-1, f(4)=-9, f '(1)=-3, f '(4)=-7, and f '' is continuous.Could you find the value of integral 1 to 4 (x f ''(x))dx. ?
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d{ xf'(x) } / dx = f'(x) + xf''(x)
∴ ∫ d{ xf'(x) }, [x=1,4] = ∫ f'(x)dx, [x=1,4] + ∫ xf''(x)dx, [x=1,4]
4f'(4) − f'(1) = f(4) − f(1) + ∫ xf''(x)dx, [x=1,4]
∴ ∫ xf''(x)dx, [x=1,4] = −28+3+9−1 = −17
∴ ∫ d{ xf'(x) }, [x=1,4] = ∫ f'(x)dx, [x=1,4] + ∫ xf''(x)dx, [x=1,4]
4f'(4) − f'(1) = f(4) − f(1) + ∫ xf''(x)dx, [x=1,4]
∴ ∫ xf''(x)dx, [x=1,4] = −28+3+9−1 = −17