已知 $\mathrm{f}(x) = ax^3 + bx^2 + cx + d$，其中 $a$、$b$、$c$ 和 $d$ 是常数。 方程为 $y = \mathrm{f}(x)$ 的曲线 $C$ 穿过点 $(0, -1)$，且在 $(-1, 1)$ 处有一个极大值点。由 $C$、 $x$ 轴以及直线 $x = 2$ 和 $x = 3$ 所围成的区域面积为 $\frac{31}{4}$ 平方单位。已知当 $2 < x < 3$ 时，$\mathrm{f}(x) > 0$，求 $a$、$b$、$c$ 和 $d$ 的值。

1. Use the coordinates of the given points to form equations involving the constants.

Since the curve passes through $(0, -1)$:
$\mathrm{f}(0) = d = -1$

Since the curve passes through $(-1, 1)$:
$\mathrm{f}(-1) = a(-1)^3 + b(-1)^2 + c(-1) - 1 = 1$
$-a + b - c - 1 = 1$
$-a + b - c = 2 \quad \text{--- (1)}$

2. Form an equation using the condition for a stationary point.

Find the derivative of $\mathrm{f}(x)$:
$\mathrm{f}'(x) = 3ax^2 + 2bx + c$

Since there is a maximum point at $x = -1$, $\mathrm{f}'(-1) = 0$:
$3a(-1)^2 + 2b(-1) + c = 0$
$3a - 2b + c = 0 \quad \text{--- (2)}$

3. Form an equation using the given area under the curve.

Given that $\mathrm{f}(x) > 0$ for $2 < x < 3$, the area is given by the definite integral:
$\int_2^3 (ax^3 + bx^2 + cx - 1) \,\mathrm{d}x = \frac{31}{4}$

Integrating and evaluating:
$\left[ \frac{a}{4}x^4 + \frac{b}{3}x^3 + \frac{c}{2}x^2 - x \right]_2^3 = \frac{31}{4}$
$\left( \frac{81a}{4} + 9b + \frac{9c}{2} - 3 \right) - \left( 4a + \frac{8b}{3} + 2c - 2 \right) = \frac{31}{4}$
$\frac{65a}{4} + \frac{19b}{3} + \frac{5c}{2} - 1 = \frac{31}{4}$
$\frac{65a}{4} + \frac{19b}{3} + \frac{5c}{2} = \frac{35}{4}$

Multiplying by $12$ to clear denominators:
$195a + 76b + 30c = 105 \quad \text{--- (3)}$

4. Solve the system of linear equations for $a$, $b$, and $c$.

From (1), $c = -a + b - 2$.
Substitute $c$ into (2):
$3a - 2b + (-a + b - 2) = 0$
$2a - b - 2 = 0 \implies b = 2a - 2$

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

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

Hence:
$b = 2(1) - 2 = 0$
$c = 1 - 4 = -3$