Answer
$f(x)=3e^{x}-5sinx+4x-2'$
Work Step by Step
Integrating $f''(x)=3e^{x}+5sinx$ with respect to x.
$f'(x)=3e^{x}-5cosx+C$; where C is an arbitary constant.
Put $f'(0)=2$
Then
$C=4$
After putting the value of C, we have
$f'(x)=3e^{x}-5cosx+4$
Now, again integrate this expression, we get
$f(x)=3e^{x}-5sinx+4x+C'$
Put $f(0)=1$
$C'=-2$
After putting the value of C', we have
Hence, $f(x)=3e^{x}-5sinx+4x-2'$
