`int_(0)^(2)(3x^(2)+2x-1)dx=`

Options

A. 7
B. 8
C. 9
D. 10

Answer

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

### Step 1: Set up the integral
Let \( I = \int_{0}^{2} (3x^{2} + 2x - 1) \, dx \).

### Step 2: Use the linearity of integration
We can split the integral into three separate integrals:
\[ I = \int_{0}^{2} 3x^{2} \, dx + \int_{0}^{2} 2x \, dx - \int_{0}^{2} 1 \, dx \]

### Step 3: Factor out constants
Now, we can factor out the constants from the integrals:
\[ I = 3 \int_{0}^{2} x^{2} \, dx + 2 \int_{0}^{2} x \, dx - \int_{0}^{2} 1 \, dx \]

### Step 4: Calculate each integral
Now, we will calculate each integral separately.

1. **For \( \int_{0}^{2} x^{2} \, dx \)**:
\[ \int x^{2} \, dx = \frac{x^{3}}{3} \quad \text{(evaluated from 0 to 2)} \]
\[ = \left[ \frac{2^{3}}{3} - \frac{0^{3}}{3} \right] = \frac{8}{3} \]

2. **For \( \int_{0}^{2} x \, dx \)**:
\[ \int x \, dx = \frac{x^{2}}{2} \quad \text{(evaluated from 0 to 2)} \]
\[ = \left[ \frac{2^{2}}{2} - \frac{0^{2}}{2} \right] = 2 \]

3. **For \( \int_{0}^{2} 1 \, dx \)**:
\[ \int 1 \, dx = x \quad \text{(evaluated from 0 to 2)} \]
\[ = [2 - 0] = 2 \]

### Step 5: Substitute back into the equation
Now substitute these results back into the equation for \( I \):
\[ I = 3 \left( \frac{8}{3} \right) + 2(2) - 2 \]
\[ = 8 + 4 - 2 \]
\[ = 10 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{2} (3x^{2} + 2x - 1) \, dx = 10 \]

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