The area between `x=y+3` and `x=y^(2)` from y = -1 to `y=1` is

Options

A. `(4)/(3)`
B. `(16)/(3)`
C. `(5)/(3)`
D. 4

Answer

To find the area between the curves \( x = y + 3 \) and \( x = y^2 \) from \( y = -1 \) to \( y = 1 \), we can follow these steps:

### Step 1: Identify the curves
The first curve is a linear equation:
\[ x = y + 3 \]
The second curve is a quadratic equation:
\[ x = y^2 \]

### Step 2: Determine the points of intersection
To find the area between the curves, we first need to find the points where they intersect. We set the equations equal to each other:
\[ y + 3 = y^2 \]
Rearranging gives us:
\[ y^2 - y - 3 = 0 \]
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = -3 \):
\[ y = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 12}}{2} = \frac{1 \pm \sqrt{13}}{2} \]
The roots are:
\[ y_1 = \frac{1 + \sqrt{13}}{2}, \quad y_2 = \frac{1 - \sqrt{13}}{2} \]
However, since we are only interested in the area from \( y = -1 \) to \( y = 1 \), we can evaluate the curves at these points.

### Step 3: Set up the integral for area
The area \( A \) between the curves from \( y = -1 \) to \( y = 1 \) can be calculated using the integral:
\[ A = \int_{-1}^{1} \left( (y + 3) - (y^2) \right) dy \]

### Step 4: Calculate the integral
Now we compute the integral:
\[ A = \int_{-1}^{1} (y + 3 - y^2) \, dy \]
Breaking it down:
\[ A = \int_{-1}^{1} (3 + y - y^2) \, dy \]

Calculating each term separately:
1. The integral of \( 3 \):
\[ \int_{-1}^{1} 3 \, dy = 3[y]_{-1}^{1} = 3(1 - (-1)) = 3 \cdot 2 = 6 \]
2. The integral of \( y \):
\[ \int_{-1}^{1} y \, dy = \left[\frac{y^2}{2}\right]_{-1}^{1} = \frac{1^2}{2} - \frac{(-1)^2}{2} = \frac{1}{2} - \frac{1}{2} = 0 \]
3. The integral of \( -y^2 \):
\[ \int_{-1}^{1} -y^2 \, dy = -\left[\frac{y^3}{3}\right]_{-1}^{1} = -\left(\frac{1^3}{3} - \frac{(-1)^3}{3}\right) = -\left(\frac{1}{3} - \frac{-1}{3}\right) = -\left(\frac{1}{3} + \frac{1}{3}\right) = -\frac{2}{3} \]

### Step 5: Combine the results
Now we combine the results from the integrals:
\[ A = 6 + 0 - \frac{2}{3} = 6 - \frac{2}{3} = \frac{18}{3} - \frac{2}{3} = \frac{16}{3} \]

### Final Answer
Thus, the area between the curves \( x = y + 3 \) and \( x = y^2 \) from \( y = -1 \) to \( y = 1 \) is:
\[ \boxed{\frac{16}{3}} \]