Question

Prove that: \[ \int_0^a f(x) \, dx = \int_0^a f(x) \, dx + \int_0^a f(2a - x) \, dx \]

Hint:

Use substitution to simplify integrals and prove relations between them.

Answer 1

Step 1: Use the substitution \( u = 2a - x \). Then, \( du = -dx \). The limits of integration change as follows: when \( x = 0 \), \( u = 2a \), and when \( x = a \), \( u = a \). Step 2: Now, the second integral becomes: \[ \int_0^a f(2a - x) \, dx = \int_{2a}^a f(u) (-du) = \int_a^{2a} f(u) \, du \] Step 3: Combining the two integrals: \[ \int_0^a f(x) \, dx + \int_0^a f(2a - x) \, dx = \int_0^a f(x) \, dx + \int_a^{2a} f(u) \, du \] Step 4: The integrals now cover the interval from 0 to \( 2a \), so: \[ \int_0^a f(x) \, dx + \int_0^a f(2a - x) \, dx = \int_0^{2a} f(x) \, dx \] Thus, the result is proved: \[ \boxed{\int_0^a f(x) \, dx = \int_0^a f(x) \, dx + \int_0^a f(2a - x) \, dx} \]