antiderivação f'(x) = x³ - x + 2 para todo x e f(1) = 2

\(f´(x)=x^3-x+2\\ A\ antiderivada\ será:\\ f(x)=\frac{x^4}{4}-\frac{x^2}{2}+2x+c\ (constante)\\ f(1)=\frac{1^4}{4}-\frac{1^2}{2}+2\times1+c\\ 2=\frac{1}{4}-\frac{1}{2}+2+c\\ c=\frac{1}{4}\\ Portanto, f(x)=\frac{x^4}{4}-\frac{x^2}{2}+2x+\frac{1}{4}\)

\[\eqalign{ & f'\left( x \right) = {x^3} - x + 2 \cr & f\left( x \right) = \int_{}^{} {f'\left( x \right)} \cr & f\left( x \right) = \int_{}^{} {{x^3} - x + 2} x + c \cr & f\left( x \right) = \dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2} + 2x + c \cr & f\left( 1 \right) = \dfrac{{{1^4}}}{4} - \dfrac{{{1^2}}}{2} + 2 + c \cr & 2 = \dfrac{{{1^4}}}{4} - \dfrac{{{1^2}}}{2} + 2 + c \cr & c = \dfrac{1}{4} \cr & \cr & f\left( x \right) = \dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2} + 2x + \dfrac{1}{4} }\]

Portanto, a antiderivada será
\(\boxed{f\left( x \right) = \dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2} + 2x + \dfrac{1}{4}}\)
.