The area of the region lying between the line `x-y+2=0` and the curve `x=sqrt(y)and y-axis`, is

Options

A. 9
B. `9//2`
C. `10//3`
D. `5//2`

Answer

To find the area of the region lying between the line \(x - y + 2 = 0\) and the curve \(x = \sqrt{y}\) (along with the y-axis), we will follow these steps:

### Step 1: Rewrite the equations
First, we need to rewrite the equations in a more usable form. The equation of the line can be rearranged to:
\[ y = x + 2 \]
The equation of the curve is already given as:
\[ x = \sqrt{y} \quad \text{or} \quad y = x^2 \]

### Step 2: Find the points of intersection
To find the area between the line and the curve, we need to determine their points of intersection. Set the equations equal to each other:
\[ x + 2 = x^2 \]
Rearranging gives:
\[ x^2 - x - 2 = 0 \]
Factoring the quadratic:
\[ (x - 2)(x + 1) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = -1 \]
Since we are interested in the region where \(x \geq 0\), we will only consider \(x = 2\).

### Step 3: Set up the integral for the area
The area \(A\) between the line and the curve from \(x = 0\) to \(x = 2\) can be expressed as:
\[ A = \int_{0}^{2} \left( (x + 2) - x^2 \right) \, dx \]

### Step 4: Evaluate the integral
Now we will evaluate the integral:
\[ A = \int_{0}^{2} (x + 2 - x^2) \, dx \]
This can be simplified to:
\[ A = \int_{0}^{2} (-x^2 + x + 2) \, dx \]

Calculating the integral:
\[ A = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{0}^{2} \]
Now substituting the limits:
\[ = \left( -\frac{2^3}{3} + \frac{2^2}{2} + 2 \cdot 2 \right) - \left( -\frac{0^3}{3} + \frac{0^2}{2} + 2 \cdot 0 \right) \]
\[ = \left( -\frac{8}{3} + 2 + 4 \right) \]
\[ = -\frac{8}{3} + 6 \]
\[ = -\frac{8}{3} + \frac{18}{3} = \frac{10}{3} \]

### Final Answer
Thus, the area of the region lying between the line and the curve is:
\[ \boxed{\frac{10}{3}} \]