A) Area bounded by `sqrt(x)+sqrt(y)=2` and coordinate axes
B) Area bounded by `y^(2)=4x` and `x^(2)=4y`
C) Area bounded by `y=4x+x^(2)` between x = 0, x = 1, X-axis
D) Area bounded by `y^(2)=4x` and `y=2x`
Arrange the above statements in the ascending order of areas.

Options

A. 1) C, B, D, A
B. 2) D, C, A, B
C. 3) A, D, B, C
D. 4) B, A, C, D

Answer

To solve the problem, we need to find the areas for each of the given statements A, B, C, and D, and then arrange them in ascending order. Let's go through each part step by step.

### A) Area bounded by \( \sqrt{x} + \sqrt{y} = 2 \) and the coordinate axes

1. **Rearranging the equation**:
\[ \sqrt{y} = 2 - \sqrt{x} \]
Squaring both sides gives:
\[ y = (2 - \sqrt{x})^2 = 4 - 4\sqrt{x} + x \]

2. **Finding the intercepts**:
- When \( x = 0 \), \( \sqrt{y} = 2 \) → \( y = 4 \)
- When \( y = 0 \), \( \sqrt{x} = 2 \) → \( x = 4 \)

3. **Setting up the area integral**:
The area under the curve from \( x = 0 \) to \( x = 4 \):
\[ A = \int_0^4 (4 - 4\sqrt{x} + x) \, dx \]

4. **Calculating the integral**:
\[ A = \left[ 4x - \frac{8}{3}x^{3/2} + \frac{1}{2}x^2 \right]_0^4 \]
Evaluating at the limits:
\[ A = \left( 16 - \frac{8}{3}(8) + 8 \right) - 0 = 16 - \frac{64}{3} + 8 = 24 - \frac{64}{3} = \frac{72 - 64}{3} = \frac{8}{3} \]

### B) Area bounded by \( y^2 = 4x \) and \( x^2 = 4y \)

1. **Finding points of intersection**:
Set \( y^2 = 4x \) and \( x^2 = 4y \):
Substitute \( y = \frac{x^2}{4} \) into \( y^2 = 4x \):
\[ \left(\frac{x^2}{4}\right)^2 = 4x \implies \frac{x^4}{16} = 4x \implies x^4 - 64x = 0 \implies x(x^3 - 64) = 0 \]
Thus, \( x = 0 \) or \( x = 4 \) (since \( x^3 = 64 \)).

2. **Setting up the area integral**:
The area can be found using:
\[ A = 2 \int_0^4 \sqrt{4x} - \frac{x^2}{4} \, dx \]

3. **Calculating the integral**:
\[ A = 2 \int_0^4 (2\sqrt{x} - \frac{x^2}{4}) \, dx = 2 \left[ \frac{4}{3}x^{3/2} - \frac{1}{12}x^3 \right]_0^4 \]
Evaluating gives:
\[ A = 2 \left( \frac{4}{3}(8) - \frac{1}{12}(64) \right) = 2 \left( \frac{32}{3} - \frac{16}{3} \right) = 2 \cdot \frac{16}{3} = \frac{32}{3} \]

### C) Area bounded by \( y = 4x + x^2 \) between \( x = 0 \) and \( x = 1 \)

1. **Setting up the area integral**:
\[ A = \int_0^1 (4x + x^2) \, dx \]

2. **Calculating the integral**:
\[ A = \left[ 2x^2 + \frac{1}{3}x^3 \right]_0^1 = \left( 2(1)^2 + \frac{1}{3}(1)^3 \right) - 0 = 2 + \frac{1}{3} = \frac{6}{3} + \frac{1}{3} = \frac{7}{3} \]

### D) Area bounded by \( y^2 = 4x \) and \( y = 2x \)

1. **Finding points of intersection**:
Set \( y^2 = 4x \) and \( y = 2x \):
\[ (2x)^2 = 4x \implies 4x^2 - 4x = 0 \implies 4x(x - 1) = 0 \]
Thus, \( x = 0 \) or \( x = 1 \).

2. **Setting up the area integral**:
The area can be found using:
\[ A = \int_0^1 (2x - \sqrt{4x}) \, dx = \int_0^1 (2x - 2\sqrt{x}) \, dx \]

3. **Calculating the integral**:
\[ A = \left[ x^2 - \frac{4}{3}x^{3/2} \right]_0^1 = (1 - \frac{4}{3}) - 0 = -\frac{1}{3} \]

### Summary of Areas
- Area A: \( \frac{8}{3} \)
- Area B: \( \frac{32}{3} \)
- Area C: \( \frac{7}{3} \)
- Area D: \( \frac{1}{3} \)

### Arranging in Ascending Order
1. Area D: \( \frac{1}{3} \)
2. Area C: \( \frac{7}{3} \)
3. Area A: \( \frac{8}{3} \)
4. Area B: \( \frac{32}{3} \)

Thus, the ascending order of the areas is:
**D < C < A < B**