Question

Prove that: `int_0^(2a) f(x)dx = int_0^a f(x)dx + int_0^a 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)dx)`

∴ `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`