The area enclosed by `[x].(x-1)leyle(2sqrtx)` from `x=0` to `2` where `[x]` is the greatest interger less than or equal to x, is equal to

Options

A. `(8sqrt2/3)+(1/2)`
B. `(8sqrt2/3)`
C. `(8sqrt2/3)-(1/2)`
D. `(8/3)-(1/sqrt2)`

Answer

To find the area enclosed by the curves defined by the greatest integer function and the function \( y = 2\sqrt{x} \) from \( x = 0 \) to \( x = 2 \), we will break down the problem step by step.

### Step 1: Identify the functions
The two functions we need to consider are:
1. \( y = [x](x - 1) \) where \( [x] \) is the greatest integer less than or equal to \( x \).
2. \( y = 2\sqrt{x} \).

### Step 2: Determine the intervals for \( [x] \)
The greatest integer function \( [x] \) will take different values depending on the interval of \( x \):
- For \( 0 \leq x < 1 \), \( [x] = 0 \) → \( y = 0 \).
- For \( 1 \leq x < 2 \), \( [x] = 1 \) → \( y = 1(x - 1) = x - 1 \).
- At \( x = 2 \), \( [x] = 2 \) → \( y = 2(2 - 1) = 2 \).

### Step 3: Find the intersection points
Next, we find the points of intersection between \( y = [x](x - 1) \) and \( y = 2\sqrt{x} \):
- For \( 0 \leq x < 1 \): \( 0 = 2\sqrt{x} \) → Intersection at \( (0, 0) \).
- For \( 1 \leq x < 2 \): Set \( x - 1 = 2\sqrt{x} \).
- Rearranging gives \( x - 1 - 2\sqrt{x} = 0 \).
- Let \( \sqrt{x} = t \) → \( t^2 - 1 - 2t = 0 \) → \( t^2 - 2t - 1 = 0 \).
- Using the quadratic formula: \( t = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2} \).
- Thus, \( x = (1 + \sqrt{2})^2 = 3 + 2\sqrt{2} \) and \( x = (1 - \sqrt{2})^2 \) is not valid since it is negative.

### Step 4: Determine the area between the curves
The area can be calculated in two parts:
1. From \( x = 0 \) to \( x = 1 \): The area is simply \( 0 \) since both functions are \( 0 \).
2. From \( x = 1 \) to \( x = 2 \): The area is given by the integral of the top function minus the bottom function.

The area \( A \) is given by:
\[ A = \int_{1}^{2} (2\sqrt{x} - (x - 1)) \, dx \]

### Step 5: Evaluate the integral
Calculating the integral:
\[ A = \int_{1}^{2} (2\sqrt{x} - x + 1) \, dx \]
\[ = \int_{1}^{2} 2\sqrt{x} \, dx - \int_{1}^{2} x \, dx + \int_{1}^{2} 1 \, dx \]
Calculating each part:
1. \( \int 2\sqrt{x} \, dx = \frac{4}{3}x^{3/2} \) evaluated from \( 1 \) to \( 2 \):
\[ = \frac{4}{3}(2^{3/2} - 1^{3/2}) = \frac{4}{3}(2\sqrt{2} - 1) \]
2. \( \int x \, dx = \frac{x^2}{2} \) evaluated from \( 1 \) to \( 2 \):
\[ = \frac{2^2}{2} - \frac{1^2}{2} = 2 - \frac{1}{2} = \frac{3}{2} \]
3. \( \int 1 \, dx = x \) evaluated from \( 1 \) to \( 2 \):
\[ = 2 - 1 = 1 \]

Putting it all together:
\[ A = \left(\frac{4}{3}(2\sqrt{2} - 1)\right) - \frac{3}{2} + 1 \]
\[ = \frac{4}{3}(2\sqrt{2} - 1) - \frac{3}{2} + 1 \]
\[ = \frac{4}{3}(2\sqrt{2} - 1) - \frac{3}{2} + \frac{2}{2} \]
\[ = \frac{4}{3}(2\sqrt{2} - 1) - \frac{1}{2} \]

### Final Result
The area enclosed by the curves from \( x = 0 \) to \( x = 2 \) is:
\[ A = \frac{4}{3}(2\sqrt{2} - 1) - \frac{1}{2} \]