Key Concepts and Formulas

• Orthocentre of a Triangle: The intersection point of the altitudes of a triangle. An altitude from a vertex is a line segment perpendicular to the opposite side.
• Slope of a Line: For two points $(x_1, y_1)$ and $(x_2, y_2)$, the slope is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
• Perpendicular Lines: Two lines with slopes $m_1$ and $m_2$ are perpendicular if $m_1 m_2 = -1$.
• Point-Slope Form of a Line: The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
• Definite Integral Property (King's Rule): $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. This property is useful for integrals where the integrand or the limits of integration have symmetry.

Step-by-Step Solution

Step 1: Find the relationship between the orthocentre coordinates $(a,b)$.

• Let the vertices of the triangle be $A=(1,2)$, $B=(2,3)$, and $C=(3,1)$.
• The orthocentre $(a,b)$ is the point where the altitudes of the triangle intersect. We can find the equation of one altitude and use the fact that $(a,b)$ lies on it.
• Let's find the altitude from vertex $C$ to the side $AB$.
  • First, calculate the slope of the side $AB$: $m_{AB} = \frac{3 - 2}{2 - 1} = \frac{1}{1} = 1$
  • The altitude from $C$ to $AB$ is perpendicular to $AB$. If $m_{h_C}$ is the slope of this altitude, then: $m_{h_C} \times m_{AB} = -1$ $m_{h_C} \times 1 = -1 \implies m_{h_C} = -1$
  • Now, find the equation of the altitude passing through $C=(3,1)$ with slope $m_{h_C} = -1$: $y - 1 = -1(x - 3)$ $y - 1 = -x + 3$ $x + y = 4$
• Since the orthocentre $(a,b)$ lies on this altitude, its coordinates must satisfy the equation: $a + b = 4$ This gives us a crucial relationship between $a$ and $b$.

Step 2: Apply King's Rule to $\mathrm{I}_1$.

• We are given the integrals: $\mathrm{I}_1=\int\limits_{\mathrm{a}}^{\mathrm{b}} x \sin \left(4 x-x^2\right) \mathrm{d} x \quad (*)$ $\mathrm{I}_2=\int\limits_{\mathrm{a}}^{\mathrm{b}} \sin \left(4 x-x^2\right) \mathrm{d} x$
• We use the King's Rule: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. In our case, $a+b=4$, so we replace $x$ with $4-x$.
• Applying this to $\mathrm{I}_1$: $\mathrm{I}_1 = \int\limits_{\mathrm{a}}^{\mathrm{b}} (4-x) \sin \left(4 (4-x)-(4-x)^2\right) \mathrm{d} x$
• Let's simplify the argument of the sine function: $4(4-x) - (4-x)^2 = (16 - 4x) - (16 - 8x + x^2)$ $= 16 - 4x - 16 + 8x - x^2$ $= 4x - x^2$
• So, the integral $\mathrm{I}_1$ becomes: $\mathrm{I}_1 = \int\limits_{\mathrm{a}}^{\mathrm{b}} (4-x) \sin \left(4 x-x^2\right) \mathrm{d} x \quad (**)$

Step 3: Combine the expressions for $\mathrm{I}_1$ to find the ratio.

• Now, we add Equation $(*)$ and Equation $(**)$: $\mathrm{I}_1 + \mathrm{I}_1 = \int\limits_{\mathrm{a}}^{\mathrm{b}} x \sin \left(4 x-x^2\right) \mathrm{d} x + \int\limits_{\mathrm{a}}^{\mathrm{b}} (4-x) \sin \left(4 x-x^2\right) \mathrm{d} x$ $2\mathrm{I}_1 = \int\limits_{\mathrm{a}}^{\mathrm{b}} \left( x \sin \left(4 x-x^2\right) + (4-x) \sin \left(4 x-x^2\right) \right) \mathrm{d} x$ $2\mathrm{I}_1 = \int\limits_{\mathrm{a}}^{\mathrm{b}} (x + 4 - x) \sin \left(4 x-x^2\right) \mathrm{d} x$ $2\mathrm{I}_1 = \int\limits_{\mathrm{a}}^{\mathrm{b}} 4 \sin \left(4 x-x^2\right) \mathrm{d} x$
• We can factor out the constant 4: $2\mathrm{I}_1 = 4 \int\limits_{\mathrm{a}}^{\mathrm{b}} \sin \left(4 x-x^2\right) \mathrm{d} x$
• Recognize that the integral on the right side is $\mathrm{I}_2$: $2\mathrm{I}_1 = 4 \mathrm{I}_2$
• Now, we can find the ratio $\frac{\mathrm{I}_1}{\mathrm{I}_2}$: $\frac{\mathrm{I}_1}{\mathrm{I}_2} = \frac{4}{2} = 2$

Step 4: Calculate the final expression.

• We need to find the value of $36 \frac{\mathrm{I}_1}{\mathrm{I}_2}$.
• Substitute the ratio we found: $36 \frac{\mathrm{I}_1}{\mathrm{I}_2} = 36 \times 2$ $36 \frac{\mathrm{I}_1}{\mathrm{I}_2} = 72$

Common Mistakes & Tips

• Orthocentre Calculation: Ensure the slopes of perpendicular lines are correctly used ($m_1 m_2 = -1$). A common error is to use $m_1 = m_2$.
• King's Rule Application: When applying $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, carefully substitute $(a+b-x)$ for every occurrence of $x$ in the integrand, including within composite functions like $\sin(g(x))$.
• Algebraic Simplification: Pay close attention to algebraic manipulations, especially when expanding squares or dealing with negative signs, to avoid errors in the argument of the sine function. The symmetry of $4x-x^2$ with respect to $x \to 4-x$ is key.

Summary

The problem requires finding the orthocentre of a given triangle to establish a relationship between the limits of integration ($a$ and $b$). This relationship ($a+b=4$) is then used to simplify the integral $\mathrm{I}_1$ by applying the King's Rule. By adding the original and transformed expressions for $\mathrm{I}_1$, we obtain a relationship between $\mathrm{I}_1$ and $\mathrm{I}_2$. Finally, the required expression $36 \frac{\mathrm{I}_1}{\mathrm{I}_2}$ is calculated using this relationship.

The final answer is \boxed{72}.