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