Let f and g be continuous functions on [0, a] such that f(x) = f(a - x) and g(x) + g(a - x) = 4. Then ∫₀ᵃ f(x)g(x) dx is equal to:

Let I = ∫₀ᵃ f(x)g(x) dx.
Since f(x) = f(a - x), we have
I = ∫₀ᵃ f(a - x)g(x) dx
Let u = a - x, then du = -dx. When x = 0, u = a; when x = a, u = 0.
Therefore,
I = ∫ₐ₀ f(u)g(a - u)(-du) = ∫₀ᵃ f(u)g(a - u) du = ∫₀ᵃ f(x)g(a - x) dx
Adding the two expressions for I, we have
2I = ∫₀ᵃ f(x)g(x) dx + ∫₀ᵃ f(x)g(a - x) dx = ∫₀ᵃ f(x)[g(x) + g(a - x)] dx
Since g(x) + g(a - x) = 4, we have
2I = ∫₀ᵃ f(x)(4) dx = 4∫₀ᵃ f(x) dx
Therefore, I = 2∫₀ᵃ f(x) dx