Question

Let f, f ' , f " be continuous functions in [0, ln2] and $f\left(0\right)=0,f^{\prime }\left(0\right)=3,f\left(ln2\right)=6,f^{\prime }\left(ln2\right)=4and{\int }_0^{In2}{e}^{-2x}f\left(x\right)dx=3,then{\int }_0^{In2}{e}^{-2x}f"\left(x\right)dx$ is?

• A. 28
• B. 13
• C. 12
• D. None of these

Answer 1

The correct option is B 13

${\int }_0^{1n2}{e}^{-2x}f\left(x\right)dx=3{\int }_0^{1n2}{e}^{-2x}f"\left(x\right)dx=e^{-2x}\int f"\left(x\right)dx+2\int e^{-2x}{f}^{\prime }\left(x\right)dx=e^{-2x}{f}^{\prime }\left(x\right)+2\left[e^{-2x}f\right(x)+2\int e^{-2x}f(x\left)dx\right]{|}_0^{1n2}=13$