Let $y = f(x)$ be a quadratic function with $f'(2) = 1$. Then the value of the integral $\int_{2-\pi}^{2+\pi} f(x) \sin\left(\frac{x-2}{2}\right) dx$ is ____.
Step-by-Step Solution
Key Concept: Recognize that f(2+t) - f(2-t) for quadratic f(x) = ax² + bx + c yields a linear expression 2t(4a+b) = 2t, allowing the integral to decompose into odd and even parts with respect to the substitution t = x - 2.
Given $f(x) = ax^2 + bx + c$ with $f'(2) = 4a + b = 1$, evaluate $I = \int_{-\pi}^{\pi} f(x)\sin\left(\frac{x-2}{2}\right)dx$ by substituting $x - 2 = t$. Using the property that $f(2+t) - f(2-t) = 8at + 2bt = 2t(4a+b) = 2t$, combine equations from $I$ and its transformation to get $2I = \int_{-\pi}^{\pi} 2t\sin\left(\frac{t}{2}\right)dt = 2\int_0^{\pi} t\sin\left(\frac{t}{2}\right)dt$. Evaluating using integration by parts with substitution $y = \frac{t}{2}$ yields $I = 8$ and $\int_0^{\pi/2} y\sin y\, dy = 1$.
Correct Answer: 8