The area of the region between `y=3x^(3)-x^(2)-10x` and `y= -x^(2)+2x` is

Options

A. 10
B. 12
C. 18
D. 24

Answer

To find the area of the region between the curves \( y = 3x^3 - x^2 - 10x \) and \( y = -x^2 + 2x \), we 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:

\[ 3x^3 - x^2 - 10x = -x^2 + 2x \]

Simplifying this, we get:

\[ 3x^3 - x^2 - 10x + x^2 - 2x = 0 \]

This simplifies to:

\[ 3x^3 - 12x = 0 \]

Factoring out \( 3x \):

\[ 3x(x^2 - 4) = 0 \]

Setting each factor to zero gives:

\[ 3x = 0 \quad \Rightarrow \quad x = 0 \]
\[ x^2 - 4 = 0 \quad \Rightarrow \quad x = \pm 2 \]

Thus, the intersection points are \( x = -2, 0, 2 \).

### Step 2: Determine the area between the curves
The area \( A \) between the curves from \( x = -2 \) to \( x = 0 \) and from \( x = 0 \) to \( x = 2 \) can be calculated using integrals.

For the interval \( [-2, 0] \):
The upper curve is \( y = 3x^3 - x^2 - 10x \) and the lower curve is \( y = -x^2 + 2x \).

The area \( A_1 \) is given by:

\[ A_1 = \int_{-2}^{0} \left( (3x^3 - x^2 - 10x) - (-x^2 + 2x) \right) dx \]

This simplifies to:

\[ A_1 = \int_{-2}^{0} (3x^3 - 10x + 2x) dx = \int_{-2}^{0} (3x^3 - 8x) dx \]

For the interval \( [0, 2] \):
The upper curve is \( y = -x^2 + 2x \) and the lower curve is \( y = 3x^3 - x^2 - 10x \).

The area \( A_2 \) is given by:

\[ A_2 = \int_{0}^{2} \left( (-x^2 + 2x) - (3x^3 - x^2 - 10x) \right) dx \]

This simplifies to:

\[ A_2 = \int_{0}^{2} (-x^2 + 2x - 3x^3 + x^2 + 10x) dx = \int_{0}^{2} (-3x^3 + 12x) dx \]

### Step 3: Calculate the integrals
Now we calculate both integrals:

1. For \( A_1 \):

\[ A_1 = \int_{-2}^{0} (3x^3 - 8x) dx \]

Calculating the integral:

\[ = \left[ \frac{3}{4}x^4 - 4x^2 \right]_{-2}^{0} \]

Evaluating at the limits:

\[ = \left( \frac{3}{4}(0)^4 - 4(0)^2 \right) - \left( \frac{3}{4}(-2)^4 - 4(-2)^2 \right) \]
\[ = 0 - \left( \frac{3}{4}(16) - 4(4) \right) \]
\[ = 0 - \left( 12 - 16 \right) = 0 + 4 = 4 \]

2. For \( A_2 \):

\[ A_2 = \int_{0}^{2} (-3x^3 + 12x) dx \]

Calculating the integral:

\[ = \left[ -\frac{3}{4}x^4 + 6x^2 \right]_{0}^{2} \]

Evaluating at the limits:

\[ = \left( -\frac{3}{4}(2)^4 + 6(2)^2 \right) - \left( -\frac{3}{4}(0)^4 + 6(0)^2 \right) \]
\[ = \left( -\frac{3}{4}(16) + 6(4) \right) - 0 \]
\[ = -12 + 24 = 12 \]

### Step 4: Total Area
The total area \( A \) is:

\[ A = A_1 + A_2 = 4 + 12 = 16 \]

### Final Result
Thus, the area of the region between the curves is:

\[ \boxed{16} \]