The area bounded by `y = f(x),` x-axis and the line `y = 1,` where `f(x)=1+1/x int_1^xf(t)dt` is

Options

A. 2 (e + 1)
B. `2 (1 - (1)/(e ))`
C. 2 (e - 1)
D. none of these

Answer

To solve the problem, we need to find the area bounded by the curve \( y = f(x) \), the x-axis, and the line \( y = 1 \), where \( f(x) = 1 + \frac{1}{x} \int_1^x f(t) dt \).

### Step-by-Step Solution:

**Step 1: Define the function \( f(x) \)**

Given:
\[ f(x) = 1 + \frac{1}{x} \int_1^x f(t) dt \]

**Step 2: Multiply both sides by \( x \)**

Multiply both sides by \( x \):
\[ x f(x) = x + \int_1^x f(t) dt \]

**Step 3: Differentiate both sides with respect to \( x \)**

Using the product rule on the left side:
\[ f(x) + x f'(x) = 1 + f(x) \]

**Step 4: Simplify the equation**

Subtract \( f(x) \) from both sides:
\[ x f'(x) = 1 \]

**Step 5: Solve for \( f'(x) \)**

\[ f'(x) = \frac{1}{x} \]

**Step 6: Integrate to find \( f(x) \)**

Integrate both sides:
\[ f(x) = \ln x + C \]

**Step 7: Determine the constant \( C \)**

Using the condition \( f(1) = 1 \):
\[ f(1) = \ln(1) + C = 0 + C = 1 \implies C = 1 \]

Thus, we have:
\[ f(x) = \ln x + 1 \]

**Step 8: Find the area bounded by \( f(x) \), the x-axis, and the line \( y = 1 \)**

Set \( f(x) = 1 \) to find the intersection point:
\[ \ln x + 1 = 1 \implies \ln x = 0 \implies x = 1 \]

**Step 9: Find the point where \( f(x) = 0 \)**

Set \( f(x) = 0 \):
\[ \ln x + 1 = 0 \implies \ln x = -1 \implies x = \frac{1}{e} \]

**Step 10: Calculate the area**

The area \( A \) is given by:
\[ A = \text{Area of rectangle} - \text{Area under the curve} \]

The rectangle area from \( x = \frac{1}{e} \) to \( x = 1 \) is:
\[ \text{Area of rectangle} = 1 \times 1 = 1 \]

Now, calculate the area under the curve from \( x = \frac{1}{e} \) to \( x = 1 \):
\[ \text{Area under the curve} = \int_{\frac{1}{e}}^1 f(x) \, dx = \int_{\frac{1}{e}}^1 (\ln x + 1) \, dx \]

**Step 11: Evaluate the integral**

Break it down:
\[ \int_{\frac{1}{e}}^1 \ln x \, dx + \int_{\frac{1}{e}}^1 1 \, dx \]

The second integral:
\[ \int_{\frac{1}{e}}^1 1 \, dx = 1 - \frac{1}{e} \]

The first integral (using integration by parts):
Let \( u = \ln x \) and \( dv = dx \):
\[ \int \ln x \, dx = x \ln x - x + C \]
Evaluating from \( \frac{1}{e} \) to \( 1 \):
\[ \left[ x \ln x - x \right]_{\frac{1}{e}}^1 = (1 \cdot 0 - 1) - \left(\frac{1}{e} \cdot (-1) - \frac{1}{e}\right) = -1 + \left(\frac{1}{e} - \frac{1}{e}\right) = -1 \]

So, the area under the curve:
\[ -1 + \left(1 - \frac{1}{e}\right) = 0 - \frac{1}{e} = -\frac{1}{e} \]

**Step 12: Calculate the total area**

Thus, the total area \( A \):
\[ A = 1 - \left(-\frac{1}{e}\right) = 1 + \frac{1}{e} \]

### Final Answer:
The area bounded by \( y = f(x) \), the x-axis, and the line \( y = 1 \) is:
\[ A = 1 - \left(-\frac{1}{e}\right) = 1 - \frac{1}{e} \]