Integrate by parts with dv=f"(x) dx, v=f'(x), u=x and du = dx.
∫udv = uv - ∫vdu = xf'(x) - ∫f'(x) dx = xf'(x) - f(x) where the terms must be evaluated at 1 and 4.
Sds G.
asked • 03/10/19Integration Question
May I ask how to do this?
Suppose that f(1)=−6, f(4)=−9, f′(1)=3, f′(4)=9, and f′′ is continuous. Find the value of ∫41 xf′′(x)dx. (upper bound=4, lower bound=1)
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