Question

Let $f,f^{\prime },f^{\prime \prime }$ be continuous in $[0,\ln 2]$ and $f\left(0\right)=0,f^{\prime }\left(0\right)=3,f\left(\ln 2\right)=6,f^{\prime }\left(\ln 2\right)=4$
$\text{and}{\int }_0^{\ln 2}{e}^{-2x}f\left(x\right)dx=3,\text{then}{\int }_0^{\ln 2}{e}^{-2x}{f}^{\prime \prime }\left(x\right)dx$
is

Answer 1

${\int }_0^{\ln 2}{e}^{-2x}{f}^{\prime \prime }\left(x\right)dx=[e^{-2x}{f}^{\prime }\left(x\right){]}_0^{\ln 2}+2{\int }_0^{\ln 2}{e}^{-2x}{f}^{\prime }\left(x\right)dx=[\frac{1}{4}{f}^{\prime }\left(\ln 2\right)-3]+2[e^{-2x}f\left(x\right){]}_0^{\ln 2}+4{\int }_0^{\ln 2}{e}^{-2x}f\left(x\right)dx=13$