Determin integers $ m,\ n\ (m>n>0)$ for which the area of the region bounded by the curve $ y\equal{}x^2\minus{}x$ and the lines $ y\equal{}mx,\ y\equal{}nx$ is $ \frac{37}{6}$.
1. **Define the area under the curve and lines:**
   Let \( S_m \) be the area of the region bounded by the parabola \( y = x^2 - x \) and the line \( y = mx \). We need to find the area between the parabola and the line from \( x = 0 \) to the point where they intersect, which is at \( x = m + 1 \).

2. **Set up the integral for the area \( S_m \):**
   \[
   S_m = \int_0^{m+1} \left( (x^2 - x) - mx \right) \, dx
   \]
   Simplify the integrand:
   \[
   S_m = \int_0^{m+1} \left( x^2 - (m+1)x \right) \, dx
   \]

3. **Evaluate the integral:**
   \[
   S_m = \int_0^{m+1} x^2 \, dx - (m+1) \int_0^{m+1} x \, dx
   \]
   Compute each integral separately:
   \[
   \int_0^{m+1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{m+1} = \frac{(m+1)^3}{3}
   \]
   \[
   \int_0^{m+1} x \, dx = \left[ \frac{x^2}{2} \right]_0^{m+1} = \frac{(m+1)^2}{2}
   \]
   Therefore:
   \[
   S_m = \frac{(m+1)^3}{3} - (m+1) \cdot \frac{(m+1)^2}{2} = \frac{(m+1)^3}{3} - \frac{(m+1)^3}{2}
   \]
   Combine the terms:
   \[
   S_m = (m+1)^3 \left( \frac{1}{3} - \frac{1}{2} \right) = (m+1)^3 \left( \frac{2 - 3}{6} \right) = -\frac{(m+1)^3}{6}
   \]
   Since the area should be positive, we take the absolute value:
   \[
   S_m = \frac{(m+1)^3}{6}
   \]

4. **Determine the area between the two lines \( y = mx \) and \( y = nx \):**
   The area between the lines is given by:
   \[
   S_m - S_n = \frac{(m+1)^3}{6} - \frac{(n+1)^3}{6}
   \]
   Given that this area is \( \frac{37}{6} \):
   \[
   \frac{(m+1)^3}{6} - \frac{(n+1)^3}{6} = \frac{37}{6}
   \]
   Multiply through by 6:
   \[
   (m+1)^3 - (n+1)^3 = 37
   \]

5. **Solve the equation \( (m+1)^3 - (n+1)^3 = 37 \):**
   Let \( a = m + 1 \) and \( b = n + 1 \). Then:
   \[
   a^3 - b^3 = 37
   \]
   Factor the left-hand side:
   \[
   (a - b)(a^2 + ab + b^2) = 37
   \]
   Since 37 is a prime number, the only integer factor pairs are \( (1, 37) \) and \( (37, 1) \). Given \( a > b \), we have:
   \[
   a - b = 1 \quad \text{and} \quad a^2 + ab + b^2 = 37
   \]

6. **Solve the system of equations:**
   From \( a - b = 1 \), we get \( a = b + 1 \). Substitute into the second equation:
   \[
   (b+1)^2 + (b+1)b + b^2 = 37
   \]
   Simplify:
   \[
   b^2 + 2b + 1 + b^2 + b + b^2 = 37
   \]
   \[
   3b^2 + 3b + 1 = 37
   \]
   \[
   3b^2 + 3b - 36 = 0
   \]
   Divide by 3:
   \[
   b^2 + b - 12 = 0
   \]
   Factor the quadratic:
   \[
   (b + 4)(b - 3) = 0
   \]
   Thus, \( b = 3 \) (since \( b = -4 \) is not valid as \( b \) must be positive).

7. **Find \( a \) and then \( m \) and \( n \):**
   \[
   a = b + 1 = 3 + 1 = 4
   \]
   Therefore:
   \[
   m = a - 1 = 4 - 1 = 3
   \]
   \[
   n = b - 1 = 3 - 1 = 2
   \]

The final answer is \( \boxed{ (m, n) = (3, 2) } \)