Find the area of the region bounded by the curve `y=x^(3),y=x+6" and "x=0`

Explanation

To find the area of the region bounded by the curves \( y = x^3 \), \( y = x + 6 \), and the line \( x = 0 \), we can follow these steps:

### Step 1: Identify the curves and the area of interest
We have two curves:
1. \( y = x^3 \) (a cubic curve)
2. \( y = x + 6 \) (a linear function)

Additionally, we have the vertical line \( x = 0 \) (the y-axis). We need to find the area between these curves from \( x = 0 \) to the point where they intersect.

### Step 2: Find the points of intersection
To find the points of intersection, we set the equations equal to each other:
\[ x^3 = x + 6 \]
Rearranging gives:
\[ x^3 - x - 6 = 0 \]
We can use the Rational Root Theorem to test for possible rational roots. Testing \( x = 2 \):
\[ 2^3 - 2 - 6 = 8 - 2 - 6 = 0 \]
Thus, \( x = 2 \) is a root.

Now, we can factor \( x^3 - x - 6 \) using synthetic division by \( x - 2 \):
\[ x^3 - x - 6 = (x - 2)(x^2 + 2x + 3) \]
The quadratic \( x^2 + 2x + 3 \) has a negative discriminant:
\[ D = 2^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8 \]
This means it has no real roots. Therefore, the only point of intersection is \( x = 2 \).

### Step 3: Determine the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 2 \) can be found using the integral:
\[ A = \int_{0}^{2} \left( (x + 6) - (x^3) \right) \, dx \]
This simplifies to:
\[ A = \int_{0}^{2} (x + 6 - x^3) \, dx \]

### Step 4: Evaluate the integral
Now we compute the integral:
\[ A = \int_{0}^{2} (x + 6 - x^3) \, dx = \int_{0}^{2} x \, dx + \int_{0}^{2} 6 \, dx - \int_{0}^{2} x^3 \, dx \]
Calculating each integral separately:
1. \( \int x \, dx = \frac{x^2}{2} \) evaluated from 0 to 2:
\[ \left[ \frac{2^2}{2} \right] - \left[ \frac{0^2}{2} \right] = 2 \]
2. \( \int 6 \, dx = 6x \) evaluated from 0 to 2:
\[ \left[ 6 \cdot 2 \right] - \left[ 6 \cdot 0 \right] = 12 \]
3. \( \int x^3 \, dx = \frac{x^4}{4} \) evaluated from 0 to 2:
\[ \left[ \frac{2^4}{4} \right] - \left[ \frac{0^4}{4} \right] = 4 \]

Putting it all together:
\[ A = 2 + 12 - 4 = 10 \]

### Final Answer
The area of the region bounded by the curves is \( 10 \) square units.