The value of the integral `int_(0)^(4){x^(3)-6x^(2)+12x-4+(x-2)cos(x-2)}dx` is equal to

Options

A. 12
B. 16
C. 0
D. 10

Answer

To solve the integral

\[ I = \int_{0}^{4} \left( x^3 - 6x^2 + 12x - 4 + (x-2)\cos(x-2) \right) dx, \]

we can break it down into two parts:

\[ I = I_1 + I_2, \]

where

\[ I_1 = \int_{0}^{4} \left( x^3 - 6x^2 + 12x - 4 \right) dx \]

and

\[ I_2 = \int_{0}^{4} (x-2) \cos(x-2) \, dx. \]

### Step 1: Calculate \( I_1 \)

We will first compute \( I_1 \):

\[ I_1 = \int_{0}^{4} \left( x^3 - 6x^2 + 12x - 4 \right) dx. \]

Using the power rule for integration, we have:

\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C. \]

Thus, we can compute each term:

1. \(\int x^3 \, dx = \frac{x^4}{4}\)
2. \(\int -6x^2 \, dx = -6 \cdot \frac{x^3}{3} = -2x^3\)
3. \(\int 12x \, dx = 12 \cdot \frac{x^2}{2} = 6x^2\)
4. \(\int -4 \, dx = -4x\)

Putting it all together:

\[ I_1 = \left[ \frac{x^4}{4} - 2x^3 + 6x^2 - 4x \right]_{0}^{4}. \]

Now, we evaluate this from 0 to 4:

\[ = \left( \frac{4^4}{4} - 2 \cdot 4^3 + 6 \cdot 4^2 - 4 \cdot 4 \right) - \left( \frac{0^4}{4} - 2 \cdot 0^3 + 6 \cdot 0^2 - 4 \cdot 0 \right). \]

Calculating the upper limit:

\[ = \left( \frac{256}{4} - 2 \cdot 64 + 6 \cdot 16 - 16 \right) \]
\[ = 64 - 128 + 96 - 16 \]
\[ = 64 + 96 - 128 - 16 = 16. \]

Thus,

\[ I_1 = 16. \]

### Step 2: Calculate \( I_2 \)

Next, we compute \( I_2 \):

\[ I_2 = \int_{0}^{4} (x-2) \cos(x-2) \, dx. \]

To evaluate this integral, we can use the substitution \( u = x - 2 \), which gives \( du = dx \). The limits change as follows:

- When \( x = 0 \), \( u = -2 \).
- When \( x = 4 \), \( u = 2 \).

Thus, we have:

\[ I_2 = \int_{-2}^{2} u \cos(u) \, du. \]

The function \( u \cos(u) \) is an odd function (since \( \cos(u) \) is even and \( u \) is odd). The integral of an odd function over a symmetric interval around zero is zero:

\[ I_2 = 0. \]

### Step 3: Combine Results

Now we can combine the results:

\[ I = I_1 + I_2 = 16 + 0 = 16. \]

Thus, the value of the integral is

\[ \boxed{16}. \]