Let f,f',f'' be continuous in [0, ln2] and f(0)=0,f'(0)=3,f(ln2)=6,f'(ln2)=4 and ∫0ln2e−2xf(x)dx=3 then ∫0ln2e−2xf''(x)dx is

∫0ln2e−2xf"(x)dx=[e−2xf'(x)]0ln2+2∫0ln2e−2xf'(x)dx=[14f'(ln2)−3]+2[e−2xf(x)+2∫e−2xf(x)dx]0ln2=13

Let f,f',f'' be continuous in [0, ln2] and f(0)=0,f'(0)=3,f(ln2)=6,f'(ln2)=4 and ∫0ln2e−2xf(x)dx=3 then ∫0ln2e−2xf''(x)dx is

∫0ln2e−2xf"(x)dx=[e−2xf'(x)]0ln2+2∫0ln2e−2xf'(x)dx=[14f'(ln2)−3]+2[e−2xf(x)+2∫e−2xf(x)dx]0ln2=13

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Let $f, f^{'}, f^{''}$ be continuous in [0, ln2] and $f (0) = 0, f^{'} (0) = 3, f (\ln 2) = 6, f^{'} (\ln 2) = 4$ and $\int_{0}^{\ln 2} e^{- 2 x} f (x) d x = 3$ then $\int_{0}^{\ln 2} e^{- 2 x} f^{''} (x) d x$ is

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$\int_{0}^{\ln 2} e^{- 2 x} f " (x) d x = {[e^{- 2 x} f' (x)]}_{0}^{\ln 2} + 2 \int_{0}^{\ln 2} e^{- 2 x} f' (x) d x$

$= [\frac{1}{4} f' (\ln 2) - 3] + 2 {[e^{- 2 x} f (x) + 2 \int e^{- 2 x} f (x) d x]}_{0}^{\ln 2} = 13$

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Let f,f',f'' be continuous in [0, ln2] and f(0)=0,f'(0)=3,f(ln2)=6,f'(ln2)=4 and ∫0ln2e−2xf(x)dx=3 then ∫0ln2e−2xf''(x)dx is

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Let $f, f^{'}, f^{''}$ be continuous in [0, ln2] and $f (0) = 0, f^{'} (0) = 3, f (\ln 2) = 6, f^{'} (\ln 2) = 4$ and $\int_{0}^{\ln 2} e^{- 2 x} f (x) d x = 3$ then $\int_{0}^{\ln 2} e^{- 2 x} f^{''} (x) d x$ is