The area enclosed between the graph of `y=x^(3)` and the lines x=0, y=1, y=8 is

Options

A. `(45)/(4)`
B. 14
C. 7
D. None of these

Answer

To find the area enclosed between the graph of \( y = x^3 \) and the lines \( x = 0 \), \( y = 1 \), and \( y = 8 \), we can follow these steps:

### Step 1: Identify the intersection points
We need to find the points where the curve \( y = x^3 \) intersects the lines \( y = 1 \) and \( y = 8 \).

1. For \( y = 1 \):
\[ 1 = x^3 \implies x = 1 \]
So, the intersection point is \( (1, 1) \).

2. For \( y = 8 \):
\[ 8 = x^3 \implies x = 2 \]
So, the intersection point is \( (2, 8) \).

### Step 2: Set up the integral
The area we want to find is between \( y = 1 \) and \( y = 8 \) from \( x = 1 \) to \( x = 2 \). We will express the area as an integral with respect to \( y \).

The equation \( y = x^3 \) can be rewritten to express \( x \) in terms of \( y \):
\[ x = y^{1/3} \]

### Step 3: Determine the area using integration
The area \( A \) can be calculated using the integral:
\[ A = \int_{y=1}^{y=8} x \, dy = \int_{1}^{8} y^{1/3} \, dy \]

### Step 4: Calculate the integral
Now we compute the integral:
\[ A = \int_{1}^{8} y^{1/3} \, dy \]
Using the power rule for integration:
\[ \int y^{n} \, dy = \frac{y^{n+1}}{n+1} + C \]
where \( n = \frac{1}{3} \):
\[ A = \left[ \frac{y^{4/3}}{4/3} \right]_{1}^{8} = \left[ \frac{3}{4} y^{4/3} \right]_{1}^{8} \]

### Step 5: Evaluate the definite integral
Now we evaluate from \( y = 1 \) to \( y = 8 \):
\[ A = \frac{3}{4} \left( 8^{4/3} - 1^{4/3} \right) \]
Calculating \( 8^{4/3} \):
\[ 8^{4/3} = (2^3)^{4/3} = 2^4 = 16 \]
Thus,
\[ A = \frac{3}{4} (16 - 1) = \frac{3}{4} \times 15 = \frac{45}{4} \]

### Final Answer
The area enclosed between the graph of \( y = x^3 \) and the lines \( x = 0 \), \( y = 1 \), and \( y = 8 \) is:
\[ \boxed{\frac{45}{4}} \text{ square units.} \]