Area bounded by `0lexle3,0leyle` minimum `(2x+2x^2+2)` is A then 12 A is

Options

A. 164
B. 170
C. 180
D. 184

Answer

To solve the problem of finding the area bounded by the curves \(y = 2x + 2\) and \(y = x^2 + 2\) within the limits \(0 \leq x \leq 3\) and \(0 \leq y\), we will follow these steps:

### Step 1: Find the points of intersection
To find the area bounded by the curves, we first need to determine where the two curves intersect. We set the equations equal to each other:

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

Subtracting \(2\) from both sides gives:

\[ 2x = x^2 \]

Rearranging this, we have:

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

Factoring out \(x\):

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

Thus, the solutions are:

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

### Step 2: Determine the area between the curves
Next, we will find the area between the curves from \(x = 0\) to \(x = 2\). The area \(A\) can be computed using the integral of the upper curve minus the lower curve:

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

This simplifies to:

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

### Step 3: Evaluate the integral
Now we will evaluate the integral:

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

Calculating at the bounds:

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

\[ = \left[ 4 - \frac{8}{3} \right] - [0] \]

\[ = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \]

### Step 4: Calculate the area for \(x\) from 2 to 3
For \(x\) from \(2\) to \(3\), the upper curve is \(y = 2x + 2\) and the lower curve is \(y = x^2 + 2\):

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

This simplifies to:

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

Evaluating this integral:

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

Calculating at the bounds:

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

\[ = \left[ 9 - 9 \right] - \left[ 4 - \frac{8}{3} \right] \]

\[ = 0 - \left[ 4 - \frac{8}{3} \right] = -\left[ \frac{12}{3} - \frac{8}{3} \right] = -\frac{4}{3} \]

### Step 5: Combine the areas
The total area \(A\) is:

\[ A = \frac{4}{3} + 0 = \frac{4}{3} \]

### Step 6: Calculate \(12A\)
Finally, we multiply the area by \(12\):

\[ 12A = 12 \times \frac{4}{3} = 16 \]

Thus, the final answer is:

\[ \boxed{16} \]