Given $f(x)= \left\{ \begin{array}{ll} -sin(x) & x \geq 0 \\ 1-x^{2} & x<0 \end{array} \right.$
such that $F(\frac{\pi}{2})=0$. Calculate $F(\pi)+F(-1)$
The given function is discontinuous at x=0 so the integral can be split into two and can be calculated separately. With this approach i am able to get the integrals as $cos(x)+c_{1}$ for x>=0 and $x-x^{3}/3+c_{2}$ for x<0. At x=$\pi/2$, i am finding the value of $c_{1}$ which comes out to be 0. So at x=0 i am finding $c_{2}$ by using $cos(x)+c_{1}=x-x^{3}/3+c_{2}$.
With $c_{1}$=0 and $c_{2}$=1, i am plugging in the values $\pi$ and -1 respectively in $F(x)$ and am getting as $\frac{-2}{3}$.
Can you please let me know if the approach to the above formulation is correct?
$ f $ is continuous at $ I= [0,+\infty) $, so it has an antiderivative at $ I $ given by $$(\forall x\ge 0)\;\;F_1(x)=\cos(x)+C_1$$
By the same, it has an antiderivative at $(-\infty,0) $ given by $$(\forall x<0)\;\; F_2(x)=x-\frac{x^3}{3}+C_2$$
$ F $ is continuous at $ 0 $, if $C_2=C_1+1$.
$ F(\frac{\pi}{2})=0 $ if $ C_1=0 $.
So, $$F(-1)+F(\pi)=-1+\frac 13+1-1=\frac{-2}{3}$$