Determine the are between `y=2x^(2)+10 and y=4x+16`

Answer

To determine the area between the curves \( y = 2x^2 + 10 \) and \( y = 4x + 16 \), we will follow these steps:

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

\[ 2x^2 + 10 = 4x + 16 \]

Rearranging gives:

\[ 2x^2 - 4x - 6 = 0 \]

Dividing the entire equation by 2 simplifies it:

\[ x^2 - 2x - 3 = 0 \]

### Step 2: Factor the quadratic equation
Next, we factor the quadratic equation:

\[ (x - 3)(x + 1) = 0 \]

Setting each factor to zero gives us the solutions:

\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

### Step 3: Find the corresponding y-values
Now we will find the y-values for the intersection points by substituting \( x = 3 \) and \( x = -1 \) into either of the original equations. We will use \( y = 4x + 16 \):

For \( x = 3 \):

\[ y = 4(3) + 16 = 12 + 16 = 28 \]

For \( x = -1 \):

\[ y = 4(-1) + 16 = -4 + 16 = 12 \]

Thus, the intersection points are \( (3, 28) \) and \( (-1, 12) \).

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = -1 \) to \( x = 3 \) is given by the integral of the upper curve minus the lower curve. The upper curve is \( y = 4x + 16 \) and the lower curve is \( y = 2x^2 + 10 \):

\[ A = \int_{-1}^{3} \left( (4x + 16) - (2x^2 + 10) \right) \, dx \]

This simplifies to:

\[ A = \int_{-1}^{3} \left( -2x^2 + 4x + 6 \right) \, dx \]

### Step 5: Calculate the integral
Now we will compute the integral:

\[ A = \int_{-1}^{3} (-2x^2 + 4x + 6) \, dx \]

Calculating the integral term by term:

\[ = \left[ -\frac{2}{3}x^3 + 2x^2 + 6x \right]_{-1}^{3} \]

### Step 6: Evaluate the definite integral
Now we evaluate the expression at the bounds:

1. For \( x = 3 \):

\[ -\frac{2}{3}(3^3) + 2(3^2) + 6(3) = -\frac{2}{3}(27) + 2(9) + 18 = -18 + 18 + 18 = 18 \]

2. For \( x = -1 \):

\[ -\frac{2}{3}(-1^3) + 2(-1^2) + 6(-1) = -\frac{2}{3}(-1) + 2(1) - 6 = \frac{2}{3} + 2 - 6 = \frac{2}{3} + \frac{6}{3} - \frac{18}{3} = -\frac{10}{3} \]

### Step 7: Calculate the area
Now we find the area:

\[ A = 18 - \left(-\frac{10}{3}\right) = 18 + \frac{10}{3} = \frac{54}{3} + \frac{10}{3} = \frac{64}{3} \]

Thus, the area between the curves is:

\[ \boxed{\frac{64}{3}} \]

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