The area (in sq. units) bounded between `y =2xln x and y =-x` from `x=e` to `x=2e` is

Options

A. `e^(e )ln (16)`
B. `e^(2)ln(16e^(3))`
C. `e^(3)ln(16)`
D. `e^(3) ln (16e^(3))`

Answer

To find the area bounded between the curves \( y = 2x \ln x \) and \( y = -x \) from \( x = e \) to \( x = 2e \), we will follow these steps:

### Step 1: Set up the integral for the area
The area \( A \) between the two curves from \( x = e \) to \( x = 2e \) can be expressed as:
\[ A = \int_{e}^{2e} \left( 2x \ln x - (-x) \right) \, dx \]
This simplifies to:
\[ A = \int_{e}^{2e} \left( 2x \ln x + x \right) \, dx \]

### Step 2: Break down the integral
Now we can separate the integral into two parts:
\[ A = \int_{e}^{2e} 2x \ln x \, dx + \int_{e}^{2e} x \, dx \]

### Step 3: Calculate the second integral
The second integral is straightforward:
\[ \int_{e}^{2e} x \, dx = \left[ \frac{x^2}{2} \right]_{e}^{2e} = \frac{(2e)^2}{2} - \frac{e^2}{2} = \frac{4e^2}{2} - \frac{e^2}{2} = \frac{3e^2}{2} \]

### Step 4: Calculate the first integral using integration by parts
For the first integral \( \int 2x \ln x \, dx \), we will use integration by parts. Let:
- \( u = \ln x \) → \( du = \frac{1}{x} \, dx \)
- \( dv = 2x \, dx \) → \( v = x^2 \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ \int 2x \ln x \, dx = x^2 \ln x - \int x^2 \cdot \frac{1}{x} \, dx = x^2 \ln x - \int x \, dx \]
Calculating the remaining integral:
\[ \int x \, dx = \frac{x^2}{2} \]
Thus,
\[ \int 2x \ln x \, dx = x^2 \ln x - \frac{x^2}{2} \]

### Step 5: Evaluate the first integral from \( e \) to \( 2e \)
Now we evaluate:
\[ \left[ x^2 \ln x - \frac{x^2}{2} \right]_{e}^{2e} \]
Calculating at the upper limit \( x = 2e \):
\[ (2e)^2 \ln(2e) - \frac{(2e)^2}{2} = 4e^2 (\ln 2 + 1) - 2e^2 = 4e^2 \ln 2 + 4e^2 - 2e^2 = 4e^2 \ln 2 + 2e^2 \]
Calculating at the lower limit \( x = e \):
\[ e^2 \ln e - \frac{e^2}{2} = e^2 - \frac{e^2}{2} = \frac{e^2}{2} \]
Thus, the first integral evaluates to:
\[ \left( 4e^2 \ln 2 + 2e^2 \right) - \frac{e^2}{2} = 4e^2 \ln 2 + 2e^2 - \frac{e^2}{2} = 4e^2 \ln 2 + \frac{4e^2}{2} - \frac{e^2}{2} = 4e^2 \ln 2 + \frac{3e^2}{2} \]

### Step 6: Combine both integrals to find the total area
Now, we combine both parts:
\[ A = \left( 4e^2 \ln 2 + \frac{3e^2}{2} \right) + \frac{3e^2}{2} = 4e^2 \ln 2 + 3e^2 \]

### Final Answer
The area bounded between the curves from \( x = e \) to \( x = 2e \) is:
\[ A = 4e^2 \ln 2 + 3e^2 \]