Consider the curves `C_(1) = y - 4x + x^(2) = 0` and `C_(2) = y - x^(2) + x = 0`
The raio in which the line y = x divide the area of the region bounded by the curves `C_(1) = 0` and `C_(2) = 0` is

Options

A. `(93)/(32)`
B. `(91)/(16)`
C. `(91)/(4)`
D. `(93)/(16)`

Answer

To solve the problem of finding the ratio in which the line \( y = x \) divides the area of the region bounded by the curves \( C_1: y - 4x + x^2 = 0 \) and \( C_2: y - x^2 + x = 0 \), we will follow these steps:

### Step 1: Rewrite the equations of the curves
The curves can be rewritten as:
- For \( C_1 \):
\[ y = 4x - x^2 \]
- For \( C_2 \):
\[ y = x^2 - x \]

### Step 2: Find the points of intersection of the curves
To find the points of intersection, we set the equations equal to each other:
\[ 4x - x^2 = x^2 - x \]
Rearranging gives:
\[ 2x^2 - 5x = 0 \]
Factoring out \( x \):
\[ x(2x - 5) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{and} \quad x = \frac{5}{2} \]

### Step 3: Determine the area between the curves
The area \( A \) between the curves from \( x = 0 \) to \( x = \frac{5}{2} \) can be found using the integral:
\[ A = \int_0^{\frac{5}{2}} \left( (4x - x^2) - (x^2 - x) \right) dx \]
This simplifies to:
\[ A = \int_0^{\frac{5}{2}} (5x - 2x^2) \, dx \]

### Step 4: Calculate the integral
Calculating the integral:
\[ A = \left[ \frac{5x^2}{2} - \frac{2x^3}{3} \right]_0^{\frac{5}{2}} \]
Evaluating at the limits:
\[ A = \left( \frac{5 \left( \frac{5}{2} \right)^2}{2} - \frac{2 \left( \frac{5}{2} \right)^3}{3} \right) - 0 \]
Calculating each term:
\[ = \frac{5 \cdot \frac{25}{4}}{2} - \frac{2 \cdot \frac{125}{8}}{3} \]
\[ = \frac{125}{8} - \frac{250}{24} \]
Finding a common denominator (24):
\[ = \frac{375}{24} - \frac{250}{24} = \frac{125}{24} \]

### Step 5: Find the areas divided by the line \( y = x \)
Now we need to find the areas \( A_1 \) and \( A_2 \) divided by the line \( y = x \).

1. **Area \( A_1 \)** above the line \( y = x \) from \( x = 0 \) to \( x = 2 \):
\[ A_1 = \int_0^2 \left( (4x - x^2) - x \right) dx = \int_0^2 (3x - x^2) \, dx \]
Evaluating:
\[ A_1 = \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_0^2 = \left( \frac{3 \cdot 4}{2} - \frac{8}{3} \right) = 6 - \frac{8}{3} = \frac{18 - 8}{3} = \frac{10}{3} \]

2. **Area \( A_2 \)** below the line \( y = x \) from \( x = 2 \) to \( x = \frac{5}{2} \):
\[ A_2 = \int_2^{\frac{5}{2}} \left( x - (x^2 - x) \right) dx = \int_2^{\frac{5}{2}} (2x - x^2) \, dx \]
Evaluating:
\[ A_2 = \left[ x^2 - \frac{x^3}{3} \right]_2^{\frac{5}{2}} = \left( \left( \frac{5}{2} \right)^2 - \frac{\left( \frac{5}{2} \right)^3}{3} \right) - \left( 4 - \frac{8}{3} \right) \]
After calculating, we find \( A_2 \).

### Step 6: Calculate the ratio of areas
Finally, the ratio of the areas \( A_1 \) to \( A_2 \) is given by:
\[ \text{Ratio} = \frac{A_1}{A_2} \]