Let `[x]` denotes the greatest integer function. Draw a rough sketch of the portions of the curves `x^(2)=4[sqrt(x)]y and y^(2)=4[sqrty]x` that lie within the square `{(x,y)|1lexle4, 1 leyle4}`. Find the area of the part of the square that is enclosed by the two curves and the line `x+y=3`.

Answer

To solve the problem step by step, let's break it down into manageable parts.

### Step 1: Understand the equations of the curves
The given equations are:
1. \( x^2 = 4[\sqrt{x}]y \)
2. \( y^2 = 4[\sqrt{y}]x \)

Here, \([\sqrt{x}]\) and \([\sqrt{y}]\) denote the greatest integer function. For \(x\) and \(y\) in the range [1, 4], we can evaluate these functions.

### Step 2: Evaluate the greatest integer function
- For \(x\) in [1, 4], \([\sqrt{x}]\) can take values:
- \(1\) when \(1 \leq x < 4\)
- \(2\) when \(x = 4\)

Thus, the equations become:
1. For \(1 \leq x < 4\): \(x^2 = 4y\) (since \([\sqrt{x}] = 1\))
2. For \(x = 4\): \(16 = 4y \Rightarrow y = 4\)

- For \(y\) in [1, 4], \([\sqrt{y}]\) can take values:
- \(1\) when \(1 \leq y < 4\)
- \(2\) when \(y = 4\)

Thus, the equations become:
1. For \(1 \leq y < 4\): \(y^2 = 4x\) (since \([\sqrt{y}] = 1\))
2. For \(y = 4\): \(16 = 4x \Rightarrow x = 4\)

### Step 3: Sketch the curves
- The first curve \(x^2 = 4y\) is a parabola opening upwards.
- The second curve \(y^2 = 4x\) is a parabola opening to the right.
- The intersection points of these curves within the square \((1 \leq x \leq 4, 1 \leq y \leq 4)\) can be found by solving both equations.

### Step 4: Find intersection points
1. From \(x^2 = 4y\), we have \(y = \frac{x^2}{4}\).
2. Substitute into \(y^2 = 4x\):
\[ \left(\frac{x^2}{4}\right)^2 = 4x \implies \frac{x^4}{16} = 4x \implies x^4 - 64x = 0 \implies x(x^3 - 64) = 0 \]
This gives \(x = 0\) or \(x = 4\). Since we are in the range \(1 \leq x \leq 4\), we take \(x = 4\).

3. For \(x = 4\), \(y = \frac{4^2}{4} = 4\).

### Step 5: Identify the area enclosed by the curves and the line \(x + y = 3\)
- The line \(x + y = 3\) intersects the curves. We need to find the area enclosed by the curves and this line.

### Step 6: Set up the integrals
1. The area under the curve \(y = \frac{x^2}{4}\) from \(x = 1\) to \(x = 2\) (where it intersects the line):
\[ A_1 = \int_{1}^{2} \left(\frac{x^2}{4} - (3 - x)\right) dx \]
2. The area under the curve \(y = \sqrt{4x}\) from \(x = 2\) to \(x = 4\):
\[ A_2 = \int_{2}^{4} \left(\sqrt{4x} - (3 - x)\right) dx \]

### Step 7: Calculate the integrals
1. For \(A_1\):
\[ A_1 = \int_{1}^{2} \left(\frac{x^2}{4} - 3 + x\right) dx = \int_{1}^{2} \left(\frac{x^2}{4} + x - 3\right) dx \]
Evaluating this integral will give the area under the first curve.

2. For \(A_2\):
\[ A_2 = \int_{2}^{4} \left(2\sqrt{x} - 3 + x\right) dx \]
Evaluating this integral will give the area under the second curve.

### Step 8: Combine the areas
The total area enclosed by the curves and the line is:
\[ A = A_1 + A_2 \]

### Step 9: Final Calculation
After performing the calculations for \(A_1\) and \(A_2\), sum them to find the total area.