If `A(n)` represent the area bounded by the curve `y=n " lnx " ("where "n in N" and " n gt1)` and x-axis from `x=1` to `x=e`, then value of `A(n)+nA(n-1)` is equal to

Options

A. `n^(2)`
B. `en^(2)`
C. `n^(2)/(e+1)`
D. `n^(2)/(e-1)`

Explanation

To find the value of \( A(n) + nA(n-1) \), we first need to compute \( A(n) \) and \( A(n-1) \).

### Step 1: Calculate \( A(n) \)

The area \( A(n) \) is given by the integral of the function \( y = n \ln x \) from \( x = 1 \) to \( x = e \).

\[ A(n) = \int_{1}^{e} n \ln x \, dx \]

Since \( n \) is a constant, we can factor it out of the integral:

\[ A(n) = n \int_{1}^{e} \ln x \, dx \]

### Step 2: Evaluate the integral \( \int \ln x \, dx \)

To evaluate \( \int \ln x \, dx \), we can use integration by parts. Let:

- \( u = \ln x \) → \( du = \frac{1}{x} \, dx \)
- \( dv = dx \) → \( v = x \)

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

Substituting the values:

\[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C \]

### Step 3: Evaluate the definite integral

Now we evaluate \( \int_{1}^{e} \ln x \, dx \):

\[ \int_{1}^{e} \ln x \, dx = \left[ x \ln x - x \right]_{1}^{e} \]

Calculating the limits:

1. At \( x = e \):
\[ e \ln e - e = e \cdot 1 - e = e - e = 0 \]

2. At \( x = 1 \):
\[ 1 \ln 1 - 1 = 1 \cdot 0 - 1 = -1 \]

Thus,

\[ \int_{1}^{e} \ln x \, dx = 0 - (-1) = 1 \]

### Step 4: Substitute back to find \( A(n) \)

Now substituting back into our expression for \( A(n) \):

\[ A(n) = n \cdot 1 = n \]

### Step 5: Calculate \( A(n-1) \)

Using the same process for \( A(n-1) \):

\[ A(n-1) = \int_{1}^{e} (n-1) \ln x \, dx = (n-1) \int_{1}^{e} \ln x \, dx = (n-1) \cdot 1 = n - 1 \]

### Step 6: Calculate \( A(n) + nA(n-1) \)

Now we can find \( A(n) + nA(n-1) \):

\[ A(n) + nA(n-1) = n + n(n-1) \]

Expanding this:

\[ = n + n^2 - n = n^2 \]

### Final Result

Thus, the value of \( A(n) + nA(n-1) \) is:

\[ \boxed{n^2} \]