A curve is given by `b yy={(sqrt(4-x^2)),0lt=x<1sqrt((3x)),1lt=xlt=3.` Find the area lying between the curve and x-axis.

Explanation

To find the area lying between the curve and the x-axis for the given piecewise function, we will break the problem into two parts based on the intervals provided.

### Step-by-Step Solution:

1. **Identify the curves and their intervals**:
- For \(0 < x < 1\), the curve is given by \(y = \sqrt{4 - x^2}\).
- For \(1 < x < 3\), the curve is given by \(y = \sqrt{3x}\).

2. **Calculate the area under the first curve** \(y = \sqrt{4 - x^2}\) from \(x = 0\) to \(x = 1\)**:
- The area under the curve can be found using the integral:
\[ A_1 = \int_{0}^{1} \sqrt{4 - x^2} \, dx \]
- This represents a quarter of a circle with radius 2. The area of a quarter circle is:
\[ A_1 = \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (2^2) = \pi \]

3. **Calculate the area under the second curve** \(y = \sqrt{3x}\) from \(x = 1\) to \(x = 3\)**:
- The area under this curve can be calculated using the integral:
\[ A_2 = \int_{1}^{3} \sqrt{3x} \, dx \]
- To solve this integral, we can use the substitution \(u = 3x\), which gives \(du = 3dx\) or \(dx = \frac{du}{3}\). Changing the limits accordingly:
- When \(x = 1\), \(u = 3\)
- When \(x = 3\), \(u = 9\)
- The integral becomes:
\[ A_2 = \int_{3}^{9} \sqrt{u} \cdot \frac{1}{3} \, du = \frac{1}{3} \cdot \frac{2}{3} u^{3/2} \bigg|_{3}^{9} \]
- Evaluating this gives:
\[ A_2 = \frac{2}{9} \left(9^{3/2} - 3^{3/2}\right) = \frac{2}{9} \left(27 - 3\sqrt{3}\right) = \frac{2}{9} (27 - 3\sqrt{3}) \]

4. **Combine the areas**:
- The total area \(A\) between the curve and the x-axis is:
\[ A = A_1 + A_2 = \pi + \frac{2}{9} (27 - 3\sqrt{3}) \]

5. **Final expression**:
- Thus, the total area lying between the curve and the x-axis is:
\[ A = \pi + \frac{54 - 6\sqrt{3}}{9} = \pi + 6 - \frac{2\sqrt{3}}{3} \]