The area of the region bounded by the curves `y=x^(3), y=(1)/(x), x=2` and x - axis (in sq. units) is

Options

A. `4-log_(e)2`
B. `(1)/(4)+log_(e)2`
C. `3-log_(e)2`
D. `(15)/(4)-log_(e)2`

Answer

To find the area of the region bounded by the curves \( y = x^3 \), \( y = \frac{1}{x} \), \( x = 2 \), and the x-axis, we will follow these steps:

### Step 1: Identify the Points of Intersection
We need to find the points where the curves \( y = x^3 \) and \( y = \frac{1}{x} \) intersect. To do this, we set the equations equal to each other:
\[ x^3 = \frac{1}{x} \]
Multiplying both sides by \( x \) (assuming \( x \neq 0 \)):
\[ x^4 = 1 \]
Taking the fourth root of both sides, we find:
\[ x = 1 \]
Thus, the curves intersect at the point \( (1, 1) \).

### Step 2: Determine the Boundaries
The area we are interested in is bounded by:
- The curve \( y = x^3 \) from \( x = 0 \) to \( x = 1 \)
- The curve \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = 2 \)
- The vertical line \( x = 2 \)
- The x-axis

### Step 3: Set Up the Integral
The area \( A \) can be calculated as the sum of two integrals:
\[ A = \int_{0}^{1} (x^3 - 0) \, dx + \int_{1}^{2} \left(\frac{1}{x} - 0\right) \, dx \]

### Step 4: Calculate the First Integral
Calculating the first integral:
\[ \int_{0}^{1} x^3 \, dx = \left[\frac{x^4}{4}\right]_{0}^{1} = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} \]

### Step 5: Calculate the Second Integral
Calculating the second integral:
\[ \int_{1}^{2} \frac{1}{x} \, dx = [\ln x]_{1}^{2} = \ln(2) - \ln(1) = \ln(2) \]

### Step 6: Combine the Results
Now we combine the results of both integrals to find the total area:
\[ A = \frac{1}{4} + \ln(2) \]

### Final Answer
Thus, the area of the region bounded by the curves is:
\[ \boxed{\frac{1}{4} + \ln(2)} \text{ square units.} \]