Step 1: Find the intersection points.
Set $x^2-3x+3=2x^2-1$. Rearranging, $x^2+3x-4=0$, which factors as $(x+4)(x-1)=0$, so $x=-4$ and $x=1$.
Step 2: Decide which curve is on top.
Test $x=0$: first curve gives $3$, second gives $-1$. So $y=x^2-3x+3$ is the upper curve between the limits.
Step 3: Set up the area integral.
$A=\displaystyle\int_{-4}^{1}\big[(x^2-3x+3)-(2x^2-1)\big]\,dx=\displaystyle\int_{-4}^{1}(-x^2-3x+4)\,dx$.
Step 4: Antiderivative.
$\displaystyle\int(-x^2-3x+4)\,dx=-\dfrac{x^3}{3}-\dfrac{3x^2}{2}+4x$.
Step 5: Evaluate at the limits.
At $x=1$: $-\dfrac13-\dfrac32+4=\dfrac{13}{6}$. At $x=-4$: $-\dfrac{-64}{3}-\dfrac{48}{2}-16=\dfrac{64}{3}-24-16=\dfrac{64}{3}-40=-\dfrac{56}{3}$.
Step 6: Subtract.
$A=\dfrac{13}{6}-\left(-\dfrac{56}{3}\right)=\dfrac{13}{6}+\dfrac{112}{6}=\dfrac{125}{6}$.
\[ \boxed{\dfrac{125}{6}} \]