So I have the curve $y=(x-4)(x-2)(x+1) = x^3 -5x^2 + 2x + 8$.
Note here $\int_{-1}^0 y dx = \frac{61}{20}$.
The question I propose to create is as follows:
A curve has the equation $y=f(x)$, where $f(x) = x^3 + ax^2 + bx + 8$, where $a, b$ are constants. The curve meets the $x$ axis at $(-1, 0)$ and the integral $\int_{-1}^0 f(x) dx = \frac{61}{20}$.
If all checks out, they should get $a=-5$ and $b=2$.
But this is not the case.
$(-1, 0)$ being a point on the curve gives $b-a = 7$.
The integral condition gives $a/3 - b/2 + 47/10 = 0$.
These two simultaneous equations do not give $a=-5$ and $b=2$.
What is going on here?
Your integral is wrong. $\int_{-1}^0f(x)dx=\frac{61}{12}$, not $\frac{61}{20}$. Therefore, the second equation should actually be $\frac{a}{3}-\frac{b}{2}=-\frac{32}{12}$, which does indeed give $a=-5$ and $b=2$.