## Calculus 8th Edition

$f(x)=3e^{x}-5sinx+4x-2'$
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, $f(x)=3e^{x}-5sinx+4x-2'$