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$If f (0) = 1, f (2) = 3, f^{'} (2) = 5, then find the value of \int_{0}^{1} x f^{''} (2 x) d x$

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Correct option is B

I1=∫01 xf′′(2x)dx. Putting t=2x, i.e. dx=dt2, we get I1=14∫02 tf′′(t)dt =14tf′(t)02−∫02 f′(t)dt (Integrating by parts) =14tf′(t)02−f(t)02=142f′(2)−f(2)+f(0)=14(10−3+1)=2

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If f(0)=1,f(2)=3,f'(2)=5, then find the value of ∫01 xf′′(2x)dx

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$If f (0) = 1, f (2) = 3, f^{'} (2) = 5, then find the value of \int_{0}^{1} x f^{''} (2 x) d x$