Given $ f(x) =\left\{\begin{matrix} \frac{2}{3+12x^2} & x \geq 0\\ \sin(2x)+\frac{2}{3} & x<0 \end{matrix}\right.$
Anti-derivative is denoted by $F(x)$ and $F(\frac{1}{2})$=$\frac{\pi}{12}$.
Evaluate $F(\frac{-\pi}{3}) + \lim_{x \to +\infty} F(x)$.
I tried to approach the above problem by finding the indefinite integral of $f(x)$ when $x=\frac{1}{2}$ and found the constant value associated with it through the given condition. Based on it, I calculated $F(\frac{-\pi}{3})$ when $x<0$ and subsequently the limit based on the piecewise continuous function.
But the evaluation of the expression is not coming correct. Can you please let me know where it is going wrong in the above approach?
$F(x)=\frac 1 3 \arctan (2x)$ for $x >0$. This cannot be used for $x<0$. For $x <0$ we get $F(x)=-\frac 1 2 \cos (2x)+\frac {2x } 3 +C$. The constant $C$ has to be evaluated using continuity of $F$ at $0$. This gives $C=\frac1 2$ so $F(x)=-\frac 1 2 \cos (2x)+\frac {2x } 3 +\frac 1 2$. I will let you take over from here.