Find the area enclosed between the curves
`y=4x^(2)` and `y=x^(2)+3`

Answer

To find the area enclosed between the curves \( y = 4x^2 \) and \( y = x^2 + 3 \), we can follow these steps:

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

\[ 4x^2 = x^2 + 3 \]

### Step 2: Simplify the Equation
Rearranging the equation gives:

\[ 4x^2 - x^2 = 3 \implies 3x^2 = 3 \]

### Step 3: Solve for \( x \)
Dividing both sides by 3:

\[ x^2 = 1 \]

Taking the square root of both sides, we find:

\[ x = \pm 1 \]

### Step 4: Find the Corresponding \( y \) Values
Now, we substitute \( x = 1 \) and \( x = -1 \) back into either of the original equations to find the \( y \) values. Using \( y = 4x^2 \):

For \( x = 1 \):

\[ y = 4(1)^2 = 4 \]

For \( x = -1 \):

\[ y = 4(-1)^2 = 4 \]

Thus, the points of intersection are \( (1, 4) \) and \( (-1, 4) \).

### Step 5: Set Up the Integral for the Area
The area \( A \) enclosed between the curves from \( x = -1 \) to \( x = 1 \) can be calculated using the integral of the upper curve minus the lower curve:

\[ A = \int_{-1}^{1} \left( (x^2 + 3) - (4x^2) \right) \, dx \]

This simplifies to:

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

### Step 6: Evaluate the Integral
Now we evaluate the integral:

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

Calculating each part:

1. For \( \int_{-1}^{1} 3 \, dx \):

\[ = 3[x]_{-1}^{1} = 3(1 - (-1)) = 3 \times 2 = 6 \]

2. For \( \int_{-1}^{1} 3x^2 \, dx \):

\[ = 3 \left[ \frac{x^3}{3} \right]_{-1}^{1} = [x^3]_{-1}^{1} = (1^3 - (-1)^3) = 1 - (-1) = 2 \]

### Step 7: Combine the Results
Putting it all together:

\[ A = 6 - 2 = 4 \]

Thus, the area enclosed between the curves \( y = 4x^2 \) and \( y = x^2 + 3 \) is:

\[ \boxed{4} \]