The area the region enclosed by `f(x)=x^(3)-10x and g(x)=6x, x ge 0, y ge 0` is

Options

A. 24
B. 39
C. 42
D. 84

Answer

To find the area of the region enclosed by the curves \( f(x) = x^3 - 10x \) and \( g(x) = 6x \) for \( x \geq 0 \) and \( y \geq 0 \), we will follow these steps:

### Step 1: Find the points of intersection
To find the points of intersection, we set \( f(x) = g(x) \):
\[ x^3 - 10x = 6x \]
This simplifies to:
\[ x^3 - 16x = 0 \]
Factoring out \( x \):
\[ x(x^2 - 16) = 0 \]
This gives us:
\[ x = 0 \quad \text{or} \quad x^2 - 16 = 0 \]
Solving for \( x \) gives:
\[ x = 0, \quad x = 4, \quad x = -4 \]
Since we are only considering \( x \geq 0 \), the points of intersection are \( x = 0 \) and \( x = 4 \).

### Step 2: Calculate the area between the curves
The area \( A \) between the curves from \( x = 0 \) to \( x = 4 \) can be found using the integral:
\[ A = \int_{0}^{4} (g(x) - f(x)) \, dx \]
Substituting \( g(x) = 6x \) and \( f(x) = x^3 - 10x \):
\[ A = \int_{0}^{4} (6x - (x^3 - 10x)) \, dx \]
This simplifies to:
\[ A = \int_{0}^{4} (6x - x^3 + 10x) \, dx = \int_{0}^{4} (16x - x^3) \, dx \]

### Step 3: Evaluate the integral
Now we compute the integral:
\[ A = \int_{0}^{4} (16x - x^3) \, dx \]
Calculating the integral:
\[ A = \left[ 8x^2 - \frac{x^4}{4} \right]_{0}^{4} \]
Evaluating at the limits:
\[ A = \left( 8(4^2) - \frac{(4^4)}{4} \right) - \left( 8(0^2) - \frac{(0^4)}{4} \right) \]
Calculating:
\[ A = \left( 8 \cdot 16 - \frac{256}{4} \right) - 0 \]
\[ A = (128 - 64) = 64 \]

### Final Result
The area of the region enclosed by the curves \( f(x) = x^3 - 10x \) and \( g(x) = 6x \) for \( x \geq 0 \) and \( y \geq 0 \) is:
\[ \boxed{64} \]