The integral value of `int_(-2)^(0) [x^(3)+3x^(2) +3x +3+ (x+1) cos (x+1)]dx` is

Options

A. 2
B. 4
C. 0
D. 8

Answer

To solve the integral

\[ I = \int_{-2}^{0} \left( x^3 + 3x^2 + 3x + 3 + (x+1) \cos(x+1) \right) dx, \]

we can follow these steps:

### Step 1: Simplify the integrand
Notice that the polynomial part \(x^3 + 3x^2 + 3x + 3\) can be rewritten using the binomial expansion. Specifically, we can recognize that:

\[ x^3 + 3x^2 + 3x + 1 = (x + 1)^3. \]

Thus, we can express the integrand as:

\[ I = \int_{-2}^{0} \left( (x + 1)^3 + 2 + (x + 1) \cos(x + 1) \right) dx. \]

### Step 2: Split the integral
Now, we can split the integral into three parts:

\[ I = \int_{-2}^{0} (x + 1)^3 \, dx + \int_{-2}^{0} 2 \, dx + \int_{-2}^{0} (x + 1) \cos(x + 1) \, dx. \]

### Step 3: Change of variables
For the third integral, let’s make the substitution \(t = x + 1\). Then \(dx = dt\), and when \(x = -2\), \(t = -1\) and when \(x = 0\), \(t = 1\). Thus, we can rewrite the integral as:

\[ I = \int_{-1}^{1} t^3 \, dt + \int_{-2}^{0} 2 \, dx + \int_{-1}^{1} t \cos(t) \, dt. \]

### Step 4: Evaluate the integrals
1. **First integral**:

\[ \int_{-1}^{1} t^3 \, dt = 0 \quad \text{(since } t^3 \text{ is an odd function)}. \]

2. **Second integral**:

\[ \int_{-2}^{0} 2 \, dx = 2 \cdot (0 - (-2)) = 2 \cdot 2 = 4. \]

3. **Third integral**:

\[ \int_{-1}^{1} t \cos(t) \, dt = 0 \quad \text{(since } t \cos(t) \text{ is also an odd function)}. \]

### Step 5: Combine the results
Now, we can combine the results:

\[ I = 0 + 4 + 0 = 4. \]

Thus, the integral value is

\[ \boxed{4}. \]