The value of `int_0^(2) (3x ^ 2 −1)dx`

Options

A. `1-cos alpha`
B. `1+cos alpha`
C. `1`
D. `cos alpha`

Explanation

To solve the integral \( \int_0^2 (3x^2 - 1) \, dx \), we will follow these steps:

### Step 1: Set up the integral
We have the integral:
\[ \int_0^2 (3x^2 - 1) \, dx \]

### Step 2: Break down the integral
We can separate the integral into two parts:
\[ \int_0^2 (3x^2 - 1) \, dx = \int_0^2 3x^2 \, dx - \int_0^2 1 \, dx \]

### Step 3: Calculate the first integral
Now, we calculate the first integral:
\[ \int_0^2 3x^2 \, dx \]
Using the power rule for integration, we have:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]
Thus,
\[ \int 3x^2 \, dx = 3 \cdot \frac{x^{3}}{3} = x^3 \]
Now we evaluate this from 0 to 2:
\[ \left[ x^3 \right]_0^2 = 2^3 - 0^3 = 8 - 0 = 8 \]

### Step 4: Calculate the second integral
Next, we calculate the second integral:
\[ \int_0^2 1 \, dx \]
This is simply:
\[ \left[ x \right]_0^2 = 2 - 0 = 2 \]

### Step 5: Combine the results
Now we combine the results from the two integrals:
\[ \int_0^2 (3x^2 - 1) \, dx = 8 - 2 = 6 \]

### Final Answer
Thus, the value of the integral \( \int_0^2 (3x^2 - 1) \, dx \) is:
\[ \boxed{6} \]