Answer
$f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{x^2}{2}+x$
Work Step by Step
$f’’(x)=x^3-2x+1,$ $f’(0)=1,$ $f(0)=0$
$f’(x)=\int(x^3-2x+1)dx$
$f’(x)=\frac{x^4}{4}-x^2+x+C_1$
Substitute in $f’(0)=1$ to solve for $C_1$
$1=0-0+0+C_1$
$C_1=1$
$f’(x)=\frac{x^4}{4}-x^2+x+1$
$f(x)=\int(\frac{x^4}{4}-x^2+x+1)dx$
$f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{x^2}{2}+x+C_2$
Substitute in $f(0)=0$ to solve for $C_2$
$0=0-0+0+C_2$
$C_2=0$
$f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{x^2}{2}+x$
