Answer
$$f\left( x \right) = - \sin x + \frac{1}{4}{e^{2x}} + x$$
Work Step by Step
$$\eqalign{
& f''\left( x \right) = \sin x + {e^{2x}} \cr
& {\text{Integrate}} \cr
& f'\left( x \right) = \int {f''\left( x \right)} dx \cr
& f'\left( x \right) = \int {\left( {\sin x + {e^{2x}}} \right)} dx \cr
& f'\left( x \right) = - \cos x + \frac{1}{2}{e^{2x}} + {C_1}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}f'\left( 0 \right) = \frac{1}{2} \cr
& \frac{1}{2} = - \cos \left( 0 \right) + \frac{1}{2}{e^{2\left( 0 \right)}} + {C_1} \cr
& \frac{1}{2} = - 1 + \frac{1}{2} + {C_1} \cr
& {C_1} = 1 \cr
& {\text{Substitute }}{C_1}{\text{ into }}\left( {\bf{1}} \right) \cr
& f'\left( x \right) = - \cos x + \frac{1}{2}{e^{2x}} + 1 \cr
& {\text{Integrate}} \cr
& f\left( x \right) = \int {f'\left( x \right)} dx \cr
& f\left( x \right) = \int {\left( { - \cos x + \frac{1}{2}{e^{2x}} + 1} \right)} \cr
& f\left( x \right) = - \sin x + \frac{1}{4}{e^{2x}} + x + {C_2}{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{Use the initial condition }}f\left( 0 \right) = \frac{1}{4} \cr
& \frac{1}{4} = - \sin \left( 0 \right) + \frac{1}{4}{e^0} + 0 + {C_2} \cr
& {C_2} = 0 \cr
& {\text{Substitute }}{C_2}{\text{ into }}\left( {\bf{2}} \right) \cr
& f\left( x \right) = - \sin x + \frac{1}{4}{e^{2x}} + x \cr} $$
