Deducing the Total Cost function $MC = (TC)'(q) = 3q^2 - 40q+220, \quad  where \quad  1 \leq q \leq 20 $

The task is to deduce a function of $TC$, if for the production of the first item we need 291 monetary units, and to write out the cost value if $q$ = 8 units.

I have found that $TC = q^3-20q^2+220q$, but i can't understand how to use that 291 monetary units for finding the cost value
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:
$$
\begin{align}
TC(8) = TVC(8) + FC &amp;= 8^3-20 \cdot 8^2 + 220 \cdot 8 + 90\\
&amp;= 1082
\end{align}
$$