If $\int_0^a f(x)dx = \int_f(2a-x) = -f(x)$, then $\int_0^a f(x)dx$ equals:
Step-by-Step Solution
Key Concept: Using the substitution property and the given functional equation $f(2a-x) = -f(x)$ to show the integral equals its own negative.
Given that $f(2a-x) = -f(x)$, we use the property of definite integrals. Let $I = \int_0^a f(x)dx$. Using the substitution $x = 2a - t$, we get $I = \int_a^0 f(2a-t)(-dt) = \int_0^a f(2a-t)dt$. Since $f(2a-t) = -f(t)$, we have $I = -\int_0^a f(t)dt = -I$, which implies $2I = 0$, so $I = 0$.
Correct Answer: 0