Question

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

Answer 1

`int_0^(2a) f(x)dx = int_0^a f(x)  dx + int_0^a f(2a - x)dx`

R.H.S. : `int_0^a f(x)dx + int_0^a f(2a - x)dx`

= I₁ + I₂    ...(i)

Consider I₂ = `int_0^a f(2a - x)dx`

Put 2a – x = t

i.e. x = 2a – t

∴ –1 dx = 1 dt

`\implies` dx = – dt

As x varies from 0 to 2a, t varies from 2a to 0

I = `int_(2a)^a f(t)(-dt)`

= `-int_(2a)^a f(t)dt`

= `int_0^(2a) f(t)dt`   ...`(int_a^b f(x)dx = -int_b^a f(x)dx)`

= `int_0^(2a) f(x)dx`  ...`(int_a^b f(x)dx = int_a^b f(t)dt)`

∴ `int_0^a f(x)dx = int_0^(2a) f(x)dx`

From equation (i)

`int_0^a f(x)dx + int_0^a f(2a - x)dx = int_0^a f(x)dx + int_0^(2a) f(x)dx`

= `int_0^(2a) f(x)dx` : L.H.S.

Thus, `int_0^(2a) f(x)dx = int_0^a f(x)dx + int_0^a f(2a - x)dx`