Fundamental Theorem Formula? [closed]

If $f'(x) = \sqrt{1 + x^3}$ and $f(1) = 0.5$, find $f(4)$

I know its the Fundamental Theorem of Calculus with the equation: $$F(b)-F(a)=\int_{a}^{b}f(x) dx$$ However, I'm not entirely sure how to solve it.

Any help?

Consider the interval $[1,4]$, so we have

\begin{align*} f(4) - f(1) &= \int_1^4 \sqrt{1+x^3} dx\\ &= x \, _2F_1\left(-\frac{1}{2},\frac{1}{3};\frac{4}{3};-x^3\right)\Big|_{x=1}^{x=4}\\ &= 4 \, _2F_1\left(-\frac{1}{2},\frac{1}{3};\frac{4}{3};-64\right)-\, _2F_1\left(-\frac{1}{2},\frac{1}{3};\frac{4}{3};-1\right)\\ &= 12.8714. \end{align*}

So we have

$$f(4) -f(1)= 12.8714 \Rightarrow f(4) -0.5 =12.8714 \Rightarrow f(4) = 13.3714.$$

I did the integration on the right by Mathematica, see this link for the method of integration; https://mathoverflow.net/questions/152180/a-definite-integral-of-hypergeometric-function-2f1.