Final Answer:
The integral of (b - a)xf(x)dx, given f(a + b - x) = f(x), evaluates to zero.
Step-by-step explanation:
Given f(a + b - x) = f(x), we need to find the integral of (b - a)xf(x)dx. To proceed, let's perform a substitution to simplify the integral. Let u = a + b - x. This implies x = a + b - u, and dx = -du. Replacing x and dx in terms of u, the integral becomes ∫(b - a)(a + b - u)f(u)(-du).
This integral can be rewritten as (-1) times the integral of (b - a)(a + b - u)f(u)du. Simplifying further, (-1) * ∫(b - a)(a + b - u)f(u)du = ∫(a + b - u)(b - a)f(u)du. Expanding this integral, we get ∫(ab + bb - au - ab + au - aa)f(u)du.
Upon simplification, the terms cancel out, leaving us with ∫(bb - aa)f(u)du. Integrating a constant multiple of a function results in the function itself multiplied by the variable, so the integral of (bb - aa)f(u)du is (bb - aa) * ∫f(u)du.
Since the integral of f(u)du represents the antiderivative of f(x) with respect to u, integrating over a range doesn't affect the result; it remains a constant value. Thus, the integral (bb - aa) * ∫f(u)du simplifies to (bb - aa) * constant, which equals zero. Therefore, the value of ∫(b - a)xf(x)dx, given f(a + b - x) = f(x), is zero.