If `int_(0)^(1) (3x^(2) + 2x+ alpha)dx =0, " then " alpha=`

Options

A. `1/2`
B. `(-1)/2`
C. 2
D. `-2`

Answer

To solve the integral equation \( \int_{0}^{1} (3x^{2} + 2x + \alpha) \, dx = 0 \), we will follow these steps:

### Step 1: Set up the integral
We start by writing the integral:
\[ \int_{0}^{1} (3x^{2} + 2x + \alpha) \, dx = 0 \]

### Step 2: Break down the integral
We can separate the integral into three parts:
\[ \int_{0}^{1} 3x^{2} \, dx + \int_{0}^{1} 2x \, dx + \int_{0}^{1} \alpha \, dx = 0 \]

### Step 3: Calculate each integral
Now we will calculate each integral individually.

1. **First integral**:
\[ \int_{0}^{1} 3x^{2} \, dx = 3 \left[ \frac{x^{3}}{3} \right]_{0}^{1} = \left[ x^{3} \right]_{0}^{1} = 1^{3} - 0^{3} = 1 \]

2. **Second integral**:
\[ \int_{0}^{1} 2x \, dx = 2 \left[ \frac{x^{2}}{2} \right]_{0}^{1} = \left[ x^{2} \right]_{0}^{1} = 1^{2} - 0^{2} = 1 \]

3. **Third integral**:
\[ \int_{0}^{1} \alpha \, dx = \alpha \left[ x \right]_{0}^{1} = \alpha (1 - 0) = \alpha \]

### Step 4: Combine the results
Now we combine the results of the integrals:
\[ 1 + 1 + \alpha = 0 \]

### Step 5: Solve for \(\alpha\)
We simplify the equation:
\[ 2 + \alpha = 0 \]
Now, solving for \(\alpha\):
\[ \alpha = -2 \]

### Final Answer
Thus, the value of \(\alpha\) is:
\[ \alpha = -2 \]

---