Area enclosed by the graph of the function `y= In^2x-1` lying in the `4^(th)` `quadrant is

Options

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

Answer

To find the area enclosed by the graph of the function \( y = \ln^2 x - 1 \) lying in the fourth quadrant, we will follow these steps:

### Step 1: Identify the function and its behavior in the fourth quadrant
The function given is \( y = \ln^2 x - 1 \). For the graph to lie in the fourth quadrant, we need \( y < 0 \) (negative) and \( x > 0 \).

### Step 2: Determine where the function is negative
Set the function less than zero:
\[ \ln^2 x - 1 < 0 \]
This simplifies to:
\[ \ln^2 x < 1 \]
Taking the square root gives:
\[ -\ln x < 1 \quad \text{and} \quad \ln x < 1 \]
This leads to:
\[ \ln x < 1 \implies x < e \]
\[ \ln x > -1 \implies x > \frac{1}{e} \]
Thus, the function is negative in the interval:
\[ \left( \frac{1}{e}, e \right) \]

### Step 3: Set up the integral for the area
The area \( A \) in the fourth quadrant can be found by integrating the function from \( \frac{1}{e} \) to \( e \):
\[ A = \int_{\frac{1}{e}}^{e} (\ln^2 x - 1) \, dx \]

### Step 4: Split the integral
We can split the integral into two parts:
\[ A = \int_{\frac{1}{e}}^{e} \ln^2 x \, dx - \int_{\frac{1}{e}}^{e} 1 \, dx \]

### Step 5: Calculate the second integral
The second integral is straightforward:
\[ \int_{\frac{1}{e}}^{e} 1 \, dx = e - \frac{1}{e} \]

### Step 6: Calculate the first integral using integration by parts
Let \( u = \ln^2 x \) and \( dv = dx \). Then, \( du = 2 \ln x \cdot \frac{1}{x} \, dx \) and \( v = x \).

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
This gives:
\[ \int \ln^2 x \, dx = x \ln^2 x - \int x \cdot 2 \ln x \cdot \frac{1}{x} \, dx = x \ln^2 x - 2 \int \ln x \, dx \]
Now, we need to calculate \( \int \ln x \, dx \):
\[ \int \ln x \, dx = x \ln x - x \]
Substituting back, we have:
\[ \int \ln^2 x \, dx = x \ln^2 x - 2(x \ln x - x) = x \ln^2 x - 2x \ln x + 2x \]

### Step 7: Evaluate the integral from \( \frac{1}{e} \) to \( e \)
Now, we evaluate:
\[ \int_{\frac{1}{e}}^{e} \ln^2 x \, dx = \left[ x \ln^2 x - 2x \ln x + 2x \right]_{\frac{1}{e}}^{e} \]
Calculating at the limits:
1. At \( x = e \):
\[ e \ln^2 e - 2e \ln e + 2e = e(1^2) - 2e(1) + 2e = e - 2e + 2e = e \]
2. At \( x = \frac{1}{e} \):
\[ \frac{1}{e} \ln^2 \frac{1}{e} - 2 \cdot \frac{1}{e} \ln \frac{1}{e} + 2 \cdot \frac{1}{e} = \frac{1}{e}(-1)^2 - 2 \cdot \frac{1}{e}(-1) + 2 \cdot \frac{1}{e} = \frac{1}{e} + \frac{2}{e} + \frac{2}{e} = \frac{5}{e} \]

### Step 8: Combine results to find the area
Now we can combine the results:
\[ \int_{\frac{1}{e}}^{e} \ln^2 x \, dx = e - \frac{5}{e} \]
Thus,
\[ A = \left( e - \frac{5}{e} \right) - \left( e - \frac{1}{e} \right) = e - \frac{5}{e} - e + \frac{1}{e} = -\frac{4}{e} \]
Since area cannot be negative, we take the absolute value:
\[ A = \frac{4}{e} \]

### Final Answer
The area enclosed by the graph of the function \( y = \ln^2 x - 1 \) lying in the fourth quadrant is:
\[ \frac{4}{e} \]