The area of the plane figure bounded by lines `y=sqrt(x), x in [0, 1], y=x^(2), x in [1, 2] and y=-x^(2)+2x+4, x in [0, 2]` is

Options

A. `10//7`
B. `10//3`
C. `3//5`
D. `4//3`

Answer

To find the area of the plane figure bounded by the curves \( y = \sqrt{x} \) for \( x \in [0, 1] \), \( y = x^2 \) for \( x \in [1, 2] \), and \( y = -x^2 + 2x + 4 \) for \( x \in [0, 2] \), we will break down the problem into manageable steps.

### Step 1: Identify the curves and their intersections

1. **Curves**:
- \( y = \sqrt{x} \) for \( x \in [0, 1] \)
- \( y = x^2 \) for \( x \in [1, 2] \)
- \( y = -x^2 + 2x + 4 \) for \( x \in [0, 2] \)

2. **Find intersections**:
- Set \( \sqrt{x} = -x^2 + 2x + 4 \) to find intersections in the interval \( [0, 2] \).
- Set \( x^2 = -x^2 + 2x + 4 \) to find intersections in the interval \( [1, 2] \).

### Step 2: Calculate the area under each curve

#### Area under \( y = \sqrt{x} \) from \( x = 0 \) to \( x = 1 \)

\[ \text{Area}_1 = \int_0^1 \sqrt{x} \, dx \]

Calculating the integral:

\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]

Evaluating from 0 to 1:

\[ \text{Area}_1 = \left[ \frac{2}{3} x^{3/2} \right]_0^1 = \frac{2}{3} (1^{3/2} - 0^{3/2}) = \frac{2}{3} \]

#### Area under \( y = x^2 \) from \( x = 1 \) to \( x = 2 \)

\[ \text{Area}_2 = \int_1^2 x^2 \, dx \]

Calculating the integral:

\[ \int x^2 \, dx = \frac{1}{3} x^3 \]

Evaluating from 1 to 2:

\[ \text{Area}_2 = \left[ \frac{1}{3} x^3 \right]_1^2 = \frac{1}{3} (2^3 - 1^3) = \frac{1}{3} (8 - 1) = \frac{7}{3} \]

#### Area under \( y = -x^2 + 2x + 4 \) from \( x = 0 \) to \( x = 2 \)

\[ \text{Area}_3 = \int_0^2 (-x^2 + 2x + 4) \, dx \]

Calculating the integral:

\[ \int (-x^2 + 2x + 4) \, dx = -\frac{1}{3} x^3 + x^2 + 4x \]

Evaluating from 0 to 2:

\[ \text{Area}_3 = \left[ -\frac{1}{3} x^3 + x^2 + 4x \right]_0^2 \]
\[ = \left(-\frac{1}{3} (2^3) + (2^2) + 4(2)\right) - \left(-\frac{1}{3} (0^3) + (0^2) + 4(0)\right) \]
\[ = \left(-\frac{8}{3} + 4 + 8\right) = \left(-\frac{8}{3} + \frac{12}{3} + \frac{24}{3}\right) = \frac{28}{3} \]

### Step 3: Combine the areas

The total area bounded by the curves is given by:

\[ \text{Total Area} = \text{Area}_3 - (\text{Area}_1 + \text{Area}_2) \]

Substituting the values:

\[ \text{Total Area} = \frac{28}{3} - \left(\frac{2}{3} + \frac{7}{3}\right) = \frac{28}{3} - \frac{9}{3} = \frac{19}{3} \]

### Final Answer

The area of the plane figure bounded by the given curves is:

\[ \boxed{\frac{19}{3}} \]