Question

Show that `int_0^a f(x)g (x)dx = 2 int_0^a f(x) dx`  if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x) = 4.

Answer 1

Let `I = int_0^a f(x) g(x) dx`

`= int_0^a f(a - x) [4 - g(a - x)] dx`

`= 4 int_0^a f(a - x) dx - int_0^a f(a - x) g (a - x)  dx`

Let a - x = t

⇒ - dx = dt

When x = 0, t = a

and x = a, t = 0

`I = -4 int_a^0 f (t) dt + int_a^0 f (t) g (t) dt`

`= 4 int_0^a f (t) dt - int_0^a f (t) g (t)  dt`

`= 4 int_0^a f (x) dx - int_0^a f (x)g (x) dx `

`= 4 int_0^a f (x) dx - I`

⇒ `2I = 4 int_0^a f (x) dx`

Hence, `I = 2 int_0^a f (x) dx`