<s>[INST] Find $F(1) = -8$, of the antiderivative $\displaystyle f(x) = x^3 + 2 \sqrt{x}$. Consider the function $\displaystyle f(x) = x^3 + 2 \sqrt{x}$. 
Let $F(x)$ be the antiderivative of $f(x)$ with $F(1) = -8.$ 
$\dfrac{x^4}{4} + \dfrac{4x^\frac{3}{2}}{3}+C$
$-8=\dfrac{(1)^4}{4} + \dfrac{4\cdot(1)^\frac{3}{2}}{3}+C$
$-8-.25-1.{\bar 3}=C$
$-9.58{\bar3} =C$
Why is this not correct?
 [/INST] Let 
$$F(x) = \int^{x} f(t) \, dt$$
where $f(x) = x^3 + 2 \sqrt{x}$. It is seen that
$$F(x) = \frac{x^{4}}{4} + \frac{4}{3} \, x^{\frac{3}{2}} + c_{0}.$$
The constant can be found by applying $F(1) = -8$ for which
\begin{align}
F(1) = -8 &= \frac{1}{4} + \frac{4}{3} + c_{0} \\
c_{0} &= - \frac{115}{12}.
\end{align}
The function $F(x)$ can now be stated as
$$F(x) = \frac{x^{4}}{4} + \frac{4}{3} \, x^{\frac{3}{2}} - \frac{115}{12}.$$
</s>