`int_(0)^(4)((y^(2)-4y+5)sin(y-2)dy)/([2y^(2)-8y+1])` is equal to

Options

A. `0`
B. `2`
C. `-2`
D. none of these

Explanation

To solve the integral

\[ \int_{0}^{4} \frac{(y^{2}-4y+5) \sin(y-2)}{(2y^{2}-8y+1)} \, dy, \]

we will follow these steps:

### Step 1: Change of Variables
Let \( t = y - 2 \). Then, we have:

- When \( y = 0 \), \( t = 0 - 2 = -2 \).
- When \( y = 4 \), \( t = 4 - 2 = 2 \).
- The differential \( dy = dt \).

Thus, the integral becomes:

\[ \int_{-2}^{2} \frac{((t+2)^{2} - 4(t+2) + 5) \sin(t)}{(2(t+2)^{2} - 8(t+2) + 1)} \, dt. \]

### Step 2: Simplifying the Numerator
Now simplify the numerator:

\[ (t+2)^{2} - 4(t+2) + 5 = t^{2} + 4t + 4 - 4t - 8 + 5 = t^{2} + 1. \]

### Step 3: Simplifying the Denominator
Now simplify the denominator:

\[ 2(t+2)^{2} - 8(t+2) + 1 = 2(t^{2} + 4t + 4) - 8t - 16 + 1 = 2t^{2} + 8t + 8 - 8t - 16 + 1 = 2t^{2} - 7. \]

### Step 4: Rewrite the Integral
Now we can rewrite the integral:

\[ \int_{-2}^{2} \frac{(t^{2} + 1) \sin(t)}{(2t^{2} - 7)} \, dt. \]

### Step 5: Check for Odd Function
Next, we need to check if the integrand is an odd function. An odd function satisfies \( f(-t) = -f(t) \).

Let:

\[ f(t) = \frac{(t^{2} + 1) \sin(t)}{(2t^{2} - 7)}. \]

Now compute \( f(-t) \):

\[ f(-t) = \frac{((-t)^{2} + 1) \sin(-t)}{(2(-t)^{2} - 7)} = \frac{(t^{2} + 1)(-\sin(t))}{(2t^{2} - 7)} = -f(t). \]

Since \( f(-t) = -f(t) \), \( f(t) \) is an odd function.

### Step 6: Evaluate the Integral
Since the integral of an odd function over a symmetric interval around zero is zero:

\[ \int_{-a}^{a} f(t) \, dt = 0. \]

Thus, we conclude:

\[ \int_{0}^{4} \frac{(y^{2}-4y+5) \sin(y-2)}{(2y^{2}-8y+1)} \, dy = 0. \]

### Final Answer
The value of the integral is:

\[ \boxed{0}. \]