Question

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 \(\int_0^a f(x)g(x)\,dx\) is equal to :

• A. \(4\int_0^a f(x)\,dx\)
• B. \(-3\int_0^a f(x)\,dx\)
• C. \(\int_0^a f(x)\,dx\)
• D. \(2\int_0^a f(x)\,dx\)

Hint:

Apply \(\int_0^a h(x)dx = \int_0^a h(a-x)dx\) to \(f(x)g(x)\), add the two forms and use \(g(x)+g(a-x)=4\).

Answer 1

The Correct Option is D

Let \(I = \int_0^a f(x)g(x)\,dx\). Using \(x \to a-x\): \(I = \int_0^a f(a-x)g(a-x)\,dx = \int_0^a f(x)g(a-x)\,dx\) (as \(f(a-x)=f(x)\)).
Adding: \(2I = \int_0^a f(x)\,[g(x) + g(a-x)]\,dx = 4\int_0^a f(x)\,dx\).
So \(I = 2\int_0^a f(x)\,dx\).
Answer: (4) \(2\int_0^a f(x)\,dx\)