## Calculus 10th Edition

$f(x)=2x-sinx+6$
Given: $f''(x)=sinx,f'(0)=1,f(0)=6$ Integrate $f''(x)$ to get $f'(x)$. $f'(x)=\int sinxdx$ $f'(x)=-cosx+C$ Solve for $C$ by setting $f'(x)=1$ and plugging in $0$ for $x$ to get the particular solution for $f'(x)$. $1=-cos(0)+C$ $1=-1+C$ $C=2$ Thus: $f'(x)=-cosx+1$ Integrate $f'(x)$ to get $f(x)$. $f(x)=\int (-cosx+2)dx$ $f(x)=2x-sinx+C$ Find the particular solution again by setting $f(x)=6$ and plugging in $0$ for $x$ to find $C$. $6=2(0)-sin(0)+C$ $6=0-0+C$ $C=6$ Thus: $f(x)=2x-sinx+6$