The area (in sq. units) of the region bounded by the parabola `y=x^2+2" and the lines " y=x+1, x=0 " and " x=3`, is

Options

A. `15/4`
B. `15/2`
C. `21/2`
D. `17/4`

Answer

To find the area of the region bounded by the parabola \( y = x^2 + 2 \), the line \( y = x + 1 \), and the vertical lines \( x = 0 \) and \( x = 3 \), we will follow these steps:

### Step 1: Determine the Points of Intersection
First, we need to find the points where the parabola intersects the line. We set the equations equal to each other:

\[ x^2 + 2 = x + 1 \]

Rearranging gives:

\[ x^2 - x + 1 = 0 \]

To find the roots, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1, b = -1, c = 1 \):

\[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} \]

Since the discriminant is negative, the parabola and the line do not intersect. Thus, we will evaluate the area between the curves from \( x = 0 \) to \( x = 3 \).

### Step 2: Set Up the Integral
The area \( A \) between the curves from \( x = 0 \) to \( x = 3 \) can be calculated using the integral:

\[ A = \int_{0}^{3} \left( (x^2 + 2) - (x + 1) \right) \, dx \]

This simplifies to:

\[ A = \int_{0}^{3} (x^2 + 2 - x - 1) \, dx = \int_{0}^{3} (x^2 - x + 1) \, dx \]

### Step 3: Calculate the Integral
Now we compute the integral:

\[ A = \int_{0}^{3} (x^2 - x + 1) \, dx \]

Calculating the integral term by term:

\[ = \left[ \frac{x^3}{3} - \frac{x^2}{2} + x \right]_{0}^{3} \]

### Step 4: Evaluate the Limits
Now we evaluate at the upper limit \( x = 3 \):

\[ = \left( \frac{3^3}{3} - \frac{3^2}{2} + 3 \right) - \left( \frac{0^3}{3} - \frac{0^2}{2} + 0 \right) \]

Calculating this gives:

\[ = \left( 9 - \frac{9}{2} + 3 \right) - 0 \]

\[ = 9 + 3 - \frac{9}{2} = 12 - \frac{9}{2} = \frac{24}{2} - \frac{9}{2} = \frac{15}{2} \]

### Final Answer
Thus, the area of the region bounded by the given curves is:

\[ \boxed{\frac{15}{2}} \]