The area (in square units) of the closed figure bounded by `x=-1,x=2and y={{:(-x^(2)+2","x le1),(2x-1","xgt1):}` and the abscissa axis, is

Options

A. `16//3`
B. `13//3`
C. `13//3`
D. `7//3`

Answer

To find the area of the closed figure bounded by the lines \(x = -1\), \(x = 2\), the curves \(y = -x^2 + 2\) for \(x \leq 1\), \(y = 2x - 1\) for \(x > 1\), and the x-axis, we can follow these steps:

### Step 1: Identify the curves and their intersections
We have two curves:
1. \(y = -x^2 + 2\) for \(x \leq 1\)
2. \(y = 2x - 1\) for \(x > 1\)

We need to find the points where these two curves intersect. To do this, we set them equal to each other:
\[ -x^2 + 2 = 2x - 1 \]
Rearranging gives:
\[ -x^2 - 2x + 3 = 0 \quad \Rightarrow \quad x^2 + 2x - 3 = 0 \]
Factoring:
\[ (x + 3)(x - 1) = 0 \]
Thus, \(x = -3\) and \(x = 1\). Since we are only interested in \(x \leq 1\) for the first curve, the relevant intersection point is at \(x = 1\).

### Step 2: Determine the area under each curve
We will calculate the area under each curve separately and then sum these areas.

#### Area under \(y = -x^2 + 2\) from \(x = -1\) to \(x = 1\)
\[ \text{Area}_1 = \int_{-1}^{1} (-x^2 + 2) \, dx \]

Calculating the integral:
\[ \int (-x^2 + 2) \, dx = -\frac{x^3}{3} + 2x \]
Evaluating from \(-1\) to \(1\):
\[ \text{Area}_1 = \left[-\frac{(1)^3}{3} + 2(1)\right] - \left[-\frac{(-1)^3}{3} + 2(-1)\right] \]
\[ = \left[-\frac{1}{3} + 2\right] - \left[\frac{1}{3} - 2\right] \]
\[ = \left[\frac{6}{3} - \frac{1}{3}\right] - \left[-\frac{1}{3} + \frac{6}{3}\right] \]
\[ = \frac{5}{3} + \frac{5}{3} = \frac{10}{3} \]

#### Area under \(y = 2x - 1\) from \(x = 1\) to \(x = 2\)
\[ \text{Area}_2 = \int_{1}^{2} (2x - 1) \, dx \]

Calculating the integral:
\[ \int (2x - 1) \, dx = x^2 - x \]
Evaluating from \(1\) to \(2\):
\[ \text{Area}_2 = \left[(2)^2 - (2)\right] - \left[(1)^2 - (1)\right] \]
\[ = [4 - 2] - [1 - 1] = 2 - 0 = 2 \]

### Step 3: Total Area
Now, we sum the two areas:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{10}{3} + 2 = \frac{10}{3} + \frac{6}{3} = \frac{16}{3} \]

Thus, the area of the closed figure bounded by the given lines and curves is:
\[ \boxed{\frac{16}{3}} \text{ square units} \]