Deducing the Total Cost function
The total cost was 291 m.u. for 1 unit given $MC(q) \equiv \frac{dC(q)}{q} = 3 q^2 - 40 q + 220$.

First, the area under the Marginal Cost gives the Total Variable Cost, $TVC(q)$. To find this area, we will integrate the Marginal Cost:

$$
\begin{align}
TVC(q) &amp;= \int_0^q\frac{dTVC(q)}{dq} dq = \int_0^q (3q^2 - 40 q + 220) dq\\
&amp;= \left. \left(q^3 -20q^2 +220q  \right) \right|_q - \left. \left(q^3 -20q^2 +220q\right)  \right|_0\\
&amp;= {q^3 -20q^2 +220q}
\end{align}
$$
Now, you have to use the additional information of 291 m. u. to find the Fixed Cost. The Total Cost equals the Variable Cost plus the Fixed Cost (which does not depend on q):
$$\begin{align}
TC(1) = TVC(1) + FC &amp;=  291\\ 
1^3 -20 \cdot 1^2 + 220 \cdot 1 + FC &amp;= 291\\
FC &amp;= 90
\end{align}
$$

Hence, for the 8 units:
$$
\begin{align}
TC(8) = TVC(8) + FC &amp;= 8^3-20 \cdot 8^2 + 220 \cdot 8 + 90\\
&amp;= 1082
\end{align}
$$