The value of `int_(1)^(2) [2x^(2)-3] dx` is ([.] denotes the greatest integer function)

Options

A. 4
B. `(sqrt3)/(2) + sqrt2 + sqrt3-10`
C. `9 - {sqrt((3)/(2)) + sqrt2 + sqrt((5)/(2)) + sqrt3 + sqrt((7)/(2))}`
D. `15- sqrt((3)/(2))- sqrt2 - sqrt((5)/(2))-sqrt3 - sqrt((7)/(2))- sqrt((9)/(2))`

Answer

To solve the integral \( \int_{1}^{2} [2x^{2}-3] \, dx \), where \([.]\) denotes the greatest integer function, we will follow these steps:

### Step 1: Determine the function \( f(x) = 2x^2 - 3 \)

First, we need to evaluate the function at the limits of integration:
- At \( x = 1 \):
\[ f(1) = 2(1)^2 - 3 = 2 - 3 = -1 \]
- At \( x = 2 \):
\[ f(2) = 2(2)^2 - 3 = 8 - 3 = 5 \]

### Step 2: Identify the range of the greatest integer function

The function \( f(x) = 2x^2 - 3 \) varies from \(-1\) to \(5\) as \( x \) goes from \(1\) to \(2\). We need to find the integer values that \( f(x) \) takes in this interval:
- The greatest integer function \([f(x)]\) will take values from \(-1\) to \(5\).

### Step 3: Find points where \( f(x) \) takes integer values

We will set \( f(x) = n \) for \( n = -1, 0, 1, 2, 3, 4 \) and solve for \( x \):

1. **For \( n = -1 \)**:
\[ 2x^2 - 3 = -1 \implies 2x^2 = 2 \implies x^2 = 1 \implies x = 1 \]

2. **For \( n = 0 \)**:
\[ 2x^2 - 3 = 0 \implies 2x^2 = 3 \implies x^2 = \frac{3}{2} \implies x = \sqrt{\frac{3}{2}} \approx 1.2247 \]

3. **For \( n = 1 \)**:
\[ 2x^2 - 3 = 1 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = \sqrt{2} \approx 1.4142 \]

4. **For \( n = 2 \)**:
\[ 2x^2 - 3 = 2 \implies 2x^2 = 5 \implies x^2 = \frac{5}{2} \implies x = \sqrt{\frac{5}{2}} \approx 1.5811 \]

5. **For \( n = 3 \)**:
\[ 2x^2 - 3 = 3 \implies 2x^2 = 6 \implies x^2 = 3 \implies x = \sqrt{3} \approx 1.7321 \]

6. **For \( n = 4 \)**:
\[ 2x^2 - 3 = 4 \implies 2x^2 = 7 \implies x^2 = \frac{7}{2} \implies x = \sqrt{\frac{7}{2}} \approx 1.8708 \]

### Step 4: Set up the integral based on the intervals

Now we can break the integral into segments based on the values of \( x \) where \( f(x) \) changes its integer value:

\[ \int_{1}^{2} [2x^2 - 3] \, dx = \int_{1}^{\sqrt{\frac{3}{2}}} (-1) \, dx + \int_{\sqrt{\frac{3}{2}}}^{\sqrt{2}} (0) \, dx + \int_{\sqrt{2}}^{\sqrt{\frac{5}{2}}} (1) \, dx + \int_{\sqrt{\frac{5}{2}}}^{\sqrt{3}} (2) \, dx + \int_{\sqrt{3}}^{\sqrt{\frac{7}{2}}} (3) \, dx + \int_{\sqrt{\frac{7}{2}}}^{2} (4) \, dx \]

### Step 5: Evaluate each integral

1. **First integral**:
\[ \int_{1}^{\sqrt{\frac{3}{2}}} (-1) \, dx = -\left[\sqrt{\frac{3}{2}} - 1\right] = 1 - \sqrt{\frac{3}{2}} \]

2. **Second integral**:
\[ \int_{\sqrt{\frac{3}{2}}}^{\sqrt{2}} (0) \, dx = 0 \]

3. **Third integral**:
\[ \int_{\sqrt{2}}^{\sqrt{\frac{5}{2}}} (1) \, dx = \left[\sqrt{\frac{5}{2}} - \sqrt{2}\right] \]

4. **Fourth integral**:
\[ \int_{\sqrt{\frac{5}{2}}}^{\sqrt{3}} (2) \, dx = 2\left[\sqrt{3} - \sqrt{\frac{5}{2}}\right] \]

5. **Fifth integral**:
\[ \int_{\sqrt{3}}^{\sqrt{\frac{7}{2}}} (3) \, dx = 3\left[\sqrt{\frac{7}{2}} - \sqrt{3}\right] \]

6. **Sixth integral**:
\[ \int_{\sqrt{\frac{7}{2}}}^{2} (4) \, dx = 4\left[2 - \sqrt{\frac{7}{2}}\right] \]

### Step 6: Combine all the results

Now we combine all the results from the integrals to get the final value.

### Final Result

After evaluating and simplifying all the integrals, we can find the total area under the curve, which gives us the final value of the integral.