The area bounded by `y = x^(2) + 3` and `y= 2x + 3` is (in sq. units)

Options

A. `(12)/(7)`
B. `(4)/(3)`
C. `(3)/(4)`
D. `(8)/(3)`

Answer

To find the area bounded by the curves \( y = x^2 + 3 \) and \( y = 2x + 3 \), we will follow these steps:

### Step 1: Find the Points of Intersection
To find the points of intersection, we set the two equations equal to each other:

\[ x^2 + 3 = 2x + 3 \]

Subtracting \( 2x + 3 \) from both sides gives:

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

Factoring out \( x \):

\[ x(x - 2) = 0 \]

This gives us the solutions:

\[ x = 0 \quad \text{or} \quad x = 2 \]

### Step 2: Find Corresponding y-values
Now we will find the corresponding \( y \)-values for these \( x \)-values:

1. For \( x = 0 \):
\[ y = 0^2 + 3 = 3 \quad \Rightarrow \quad (0, 3) \]

2. For \( x = 2 \):
\[ y = 2^2 + 3 = 4 + 3 = 7 \quad \Rightarrow \quad (2, 7) \]

The points of intersection are \( (0, 3) \) and \( (2, 7) \).

### Step 3: Determine the Area Between the Curves
To find the area between the curves from \( x = 0 \) to \( x = 2 \), we need to identify which curve is on top.

- The curve \( y = 2x + 3 \) is a straight line with a positive slope.
- The curve \( y = x^2 + 3 \) is a parabola opening upwards.

At \( x = 0 \):
- \( y = 2(0) + 3 = 3 \)
- \( y = 0^2 + 3 = 3 \)

At \( x = 2 \):
- \( y = 2(2) + 3 = 7 \)
- \( y = 2^2 + 3 = 7 \)

For \( 0 < x < 2 \), \( y = 2x + 3 \) is above \( y = x^2 + 3 \).

### Step 4: Set Up the Integral
The area \( A \) between the curves is given by:

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

This simplifies to:

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

### Step 5: Evaluate the Integral
Now we will evaluate the integral:

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

Calculating the integral:

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

Calculating at the limits:

1. At \( x = 2 \):
\[ = 2^2 - \frac{2^3}{3} = 4 - \frac{8}{3} = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \]

2. At \( x = 0 \):
\[ = 0^2 - \frac{0^3}{3} = 0 \]

Thus, the area is:

\[ A = \frac{4}{3} \text{ square units} \]

### Final Answer
The area bounded by the curves \( y = x^2 + 3 \) and \( y = 2x + 3 \) is \( \frac{4}{3} \) square units.

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