Area bounded by the curves `(y)/(x)=log x and (y)/(2)=-x^(2)+x` (in sq. units) equals

Options

A. `(7)/(12)`
B. `(12)/(7)`
C. `(7)/(6)`
D. `(6)/(7)`

Answer

To find the area bounded by the curves \( \frac{y}{x} = \log x \) and \( \frac{y}{2} = -x^2 + x \), we will follow these steps:

### Step 1: Rewrite the equations of the curves
The first curve can be rewritten as:
\[ y = x \log x \]
The second curve can be rewritten as:
\[ y = -2x^2 + 2x \]

### Step 2: Find the points of intersection
To find the area between the curves, we need to determine the points where they intersect. We set the equations equal to each other:
\[ x \log x = -2x^2 + 2x \]
Rearranging gives:
\[ x \log x + 2x^2 - 2x = 0 \]
Factoring out \( x \):
\[ x(\log x + 2x - 2) = 0 \]
This gives us one solution at \( x = 0 \). For the other solutions, we solve:
\[ \log x + 2x - 2 = 0 \]
Using numerical methods or graphing, we find that \( x = 1 \) is another solution.

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) is given by:
\[ A = \int_{0}^{1} \left( x \log x - (-2x^2 + 2x) \right) \, dx \]
This simplifies to:
\[ A = \int_{0}^{1} \left( x \log x + 2x^2 - 2x \right) \, dx \]

### Step 4: Evaluate the integral
We can break this integral into three parts:
\[ A = \int_{0}^{1} x \log x \, dx + \int_{0}^{1} 2x^2 \, dx - \int_{0}^{1} 2x \, dx \]

1. **Integral of \( x \log x \)**:
Using integration by parts, let \( u = \log x \) and \( dv = x \, dx \):
\[ du = \frac{1}{x} \, dx, \quad v = \frac{x^2}{2} \]
Then,
\[ \int x \log x \, dx = \frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \log x - \frac{1}{4} x^2 + C \]
Evaluating from 0 to 1:
\[ \left[ \frac{1}{2} \cdot 1 \cdot \log 1 - \frac{1}{4} \cdot 1^2 \right] - \lim_{x \to 0} \left( \frac{x^2}{2} \log x - \frac{1}{4} x^2 \right) = 0 - \left( 0 \right) = -\frac{1}{4} \]

2. **Integral of \( 2x^2 \)**:
\[ \int 2x^2 \, dx = \frac{2}{3} x^3 \Big|_0^1 = \frac{2}{3} \]

3. **Integral of \( 2x \)**:
\[ \int 2x \, dx = x^2 \Big|_0^1 = 1 \]

### Step 5: Combine the results
Putting it all together:
\[ A = \left(-\frac{1}{4}\right) + \frac{2}{3} - 1 \]
Finding a common denominator (12):
\[ A = -\frac{3}{12} + \frac{8}{12} - \frac{12}{12} = \frac{8 - 3 - 12}{12} = \frac{-7}{12} \]
Since area cannot be negative, we take the absolute value:
\[ A = \frac{7}{12} \text{ square units} \]

### Final Answer
The area bounded by the curves is \( \frac{7}{12} \) square units.

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