What is \( \int_0^4 \frac{f(4-x)}{g(4-x)} dx \) equal to ?

For the next two (02) items that follow : Let \( f(x) = \sin x \) and \( g(x) - f(x) = f(4 - x) \).

  1. A. 0
  2. B. 1
  3. C. 2
  4. D. 4

Correct Answer: C. 2

Explanation

Note that \( g(4-x) = f(4-x) + f(4-(4-x)) = f(4-x) + f(x) = g(x) \). The integral becomes \( \int_0^4 \frac{f(4-x)}{g(x)} dx \), which is the transformed integral obtained by King's rule in the previous question. It is also equal to 2.

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