Let `f(x)=x^(2//3), x ge 0`. Then the area of the region enclosed by the curve `y=f(x)` and the three lines `y=x, x=1 and x=8` is

Options

A. `(63)/(2)`
B. `(93)/(5)`
C. `(105)/(7)`
D. `(129)/(10)`

Answer

To find the area of the region enclosed by the curve \( y = f(x) = x^{2/3} \) and the lines \( y = x \), \( x = 1 \), and \( x = 8 \), we will follow these steps:

### Step 1: Find the Points of Intersection
We need to find the points where the curve \( y = x^{2/3} \) intersects the line \( y = x \). To do this, we set the equations equal to each other:

\[ x^{2/3} = x \]

Rearranging gives:

\[ x^{2/3} - x = 0 \]

Factoring out \( x^{2/3} \):

\[ x^{2/3}(1 - x^{1/3}) = 0 \]

This gives us two solutions:

1. \( x^{2/3} = 0 \) → \( x = 0 \)
2. \( 1 - x^{1/3} = 0 \) → \( x^{1/3} = 1 \) → \( x = 1 \)

Thus, the points of intersection are \( (0, 0) \) and \( (1, 1) \).

### Step 2: Set Up the Integral for the Area
The area \( A \) between the curve and the line from \( x = 1 \) to \( x = 8 \) can be found using the integral:

\[ A = \int_{1}^{8} (x - x^{2/3}) \, dx \]

### Step 3: Compute the Integral
Now, we compute the integral:

\[ A = \int_{1}^{8} (x - x^{2/3}) \, dx = \int_{1}^{8} x \, dx - \int_{1}^{8} x^{2/3} \, dx \]

Calculating the first integral:

\[ \int x \, dx = \frac{x^2}{2} \]

Evaluating from 1 to 8:

\[ \left[ \frac{x^2}{2} \right]_{1}^{8} = \frac{8^2}{2} - \frac{1^2}{2} = \frac{64}{2} - \frac{1}{2} = 32 - 0.5 = 31.5 \]

Now, calculating the second integral:

\[ \int x^{2/3} \, dx = \frac{x^{5/3}}{5/3} = \frac{3}{5} x^{5/3} \]

Evaluating from 1 to 8:

\[ \left[ \frac{3}{5} x^{5/3} \right]_{1}^{8} = \frac{3}{5} (8^{5/3}) - \frac{3}{5} (1^{5/3}) = \frac{3}{5} (32) - \frac{3}{5} (1) = \frac{96}{5} - \frac{3}{5} = \frac{93}{5} \]

### Step 4: Combine the Results
Now we combine the results of both integrals:

\[ A = 31.5 - \frac{93}{5} \]

Converting \( 31.5 \) to a fraction:

\[ 31.5 = \frac{63}{2} \]

Finding a common denominator (10):

\[ \frac{63}{2} = \frac{315}{10}, \quad \frac{93}{5} = \frac{186}{10} \]

Thus,

\[ A = \frac{315}{10} - \frac{186}{10} = \frac{129}{10} \]

### Final Answer
The area of the region enclosed by the curve and the lines is:

\[ \boxed{\frac{129}{10}} \text{ square units.} \]