Integral giving wrong values Given following function:
$x^3 + 1$

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Find the area that is selected in red lines.
So I solved the root $\sqrt[3]{-1} = -1$ so $x = -1$
So now I have to create the 3 integrals:
$S_1 = \int_{-2}^{-1}\begin{pmatrix}0 - (x^3 + 1)\end{pmatrix}dx = \begin{bmatrix} \dfrac{-x^4}{4} - x\end{bmatrix}_{-2}^{-1}\textrm{}$
$\begin{bmatrix}
\dfrac{-(-2)^4}{4} + 2
\end{bmatrix} = -2$
$\begin{bmatrix}
\dfrac{-(-1)^4}{4} + 1
\end{bmatrix} = \dfrac{3}{4}$
$-2 + \dfrac{3}{4} = -\dfrac{3}{4}$
$ => S_1 = -1\dfrac{1}{4}$
$S_2 = \int_{-1}^{0}\begin{pmatrix}x^3 + 1\end{pmatrix}dx = \begin{bmatrix} \dfrac{x^4}{4} + x\end{bmatrix}_{-1}^{0}\textrm{}$
$\begin{bmatrix}
\dfrac{(-1)^4}{4} - 1
\end{bmatrix} = -1\dfrac{3}{4}$
$=> S_2 = -\dfrac{3}{4}$
$S_3 = \int_{0}^{3}\begin{pmatrix}x^3 + 1\end{pmatrix}dx = \begin{bmatrix} \dfrac{x^4}{4} + x\end{bmatrix}_{0}^{3}\textrm{}$
$\begin{bmatrix}
\dfrac{3^4}{4} + 3
\end{bmatrix} = 23\dfrac{1}{4}$
$=> S_3 = 23\dfrac{1}{4}$$
$S_1 + S_2 + S_3 = -1\dfrac{1}{4} - \dfrac{3}{4} + 23\dfrac{1}{4}  = 21\dfrac{1}{4}$
And the correct answer is $26\dfrac{3}{4}$
What did I do wrong? I've checked myself many times, but I am failing to see my mistake... thanks!

Hint: If you have $\begin{bmatrix} -\frac{x^4}{4}-x \end{bmatrix}_a^b$, then you have to calculate $-\frac{b^4}{4}-b-\left(-\frac{a^4}{4}-a \right)=-\frac{b^4}{4}-b+\frac{a^4}{4}+a$
You have to substract the whole term, if you insert the lower bound. And both signs change, if you remove the brackets.
Additionally you have to take the absolute value of the subareas. 
