It is given that $f(x)=ax^3+bx^2+cx+d$, where $a$, $b$, $c$ and $d$ are constants. The curve $C$ with equation $y=f(x)$ passes through $(0,-1)$ and has a maximum point at $(-1,1)$. The area bounded by $C$, the x-axis and the lines $x=2$ and $x=3$ is $\frac{31}{4}$ units^2. Given that $f(x)>0$ for $2<x<3$, find the values of $a$, $b$, $c$ and $d.

1. Determine the value of $d$ using the given y-intercept.

Given $f(x) = ax^3 + bx^2 + cx + d$.
Since the curve passes through $(0, -1)$:
$f(0) = a(0)^3 + b(0)^2 + c(0) + d = -1$
$d = -1$

2. Form equations involving $a$, $b$, and $c$ using the coordinates and properties of the maximum point.

The curve passes through $(-1, 1)$:

$-a + b - c = 2 \quad \text{--- (1)}$

Since there is a maximum point at $x = -1$, the first derivative is zero ($f'(-1) = 0$).
$f'(x) = 3ax^2 + 2bx + c$

$3a - 2b + c = 0 \quad \text{--- (2)}$

3. Set up a third equation using the given area under the curve.

The area bounded by $C$, the x-axis, $x=2$ and $x=3$ is $\frac{31}{4}$ units$^2$. Since $f(x) > 0$ for $2 < x < 3$:
$\int_{2}^{3} (ax^3 + bx^2 + cx - 1) \,dx = \frac{31}{4}$
$\left[ \frac{a}{4}x^4 + \frac{b}{3}x^3 + \frac{c}{2}x^2 - x \right]_{2}^{3} = \frac{31}{4}$
$\left( \frac{81a}{4} + 9b + \frac{9c}{2} - 3 \right) - \left( 4a + \frac{8b}{3} + 2c - 2 \right) = \frac{31}{4}$
$\frac{65}{4}a + \frac{19}{3}b + \frac{5}{2}c - 1 = \frac{31}{4}$
$\frac{65}{4}a + \frac{19}{3}b + \frac{5}{2}c = \frac{35}{4}$
Multiplying by $12$ to clear the denominators:
$195a + 76b + 30c = 105 \quad \text{--- (3)}$

4. Solve the system of three linear equations simultaneously to find $a$, $b$, and $c$.

Solving equations (1), (2) and (3) simultaneously.
Adding (1) and (2):
$(-a+b-c) + (3a-2b+c) = 2 + 0$
$2a - b = 2 \implies b = 2a - 2$

Substitute $b$ into (1):
$-a + (2a - 2) - c = 2 \implies c = a - 4$

Substitute expressions for $b$ and $c$ into (3):
$195a + 76(2a - 2) + 30(a - 4) = 105$
$195a + 152a - 152 + 30a - 120 = 105$
$377a - 272 = 105$
$377a = 377 \implies a = 1$

Substitute $a=1$ back to find $b$ and $c$:
$b = 2(1) - 2 = 0$
$c = 1 - 4 = -3$

(Alternatively, solving the system of linear equations directly via a Graphing Calculator yields $a=1, b=0, c=-3$.)