Question

Prove that: `int_0^(2"a") "f"(x)  "d"x = int_0^"a" "f"(x)  "d"x + int_0^"a" "f"(2"a" - x)  "d"x`

Answer 1

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

Let I = `int_0^"a""f"(x)"d"x + int_0^"a" "f"(2"a" - x)"d"x`

= I₁ + I₂    ........(i)

Consider I₂ = `int_0^"a" "f"(2"a" - x)  "d"x`

Put 2a – x = t

∴ − dx = dt

∴ dx = – dt

When x = 0, t = 2a – 0 = 2a

and when x = a, t = 2a – a = a

= I₂ = `int_(2"a")^"a" "f"("t")(- "dt")`

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

= `-int_"a"^(2"a") "f"("t") "dt"`   ......`[∵ int_"a"^"b" "f"(x)  "d"x = -int_"b"^"a" "f"(x)  "d"x]`

= `-int_"a"^(2"a") "f"(x) "d"x`   ......`[∵ int_"a"^"b" "f"(x)  "d"x = -int_"a"^"b" "f"("t")  "d"x]`

From (i), I = I₁ + I₂

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

= `int_0^"a" "f"(x)  "d"x + int_"a"^(2"a") "f"(x)  "d"x`

= `int_0^(2"a") "f"(x)  "d"x`  .......`[∵ int_"a"^"b" "f"(x) "d"x = int_"a"^"c" "f"(x) "d"x + int_"c"^"b" "f"(x)  "d"x; "a" < "c" < "b"]`

= L.H.S

∴ `int_0^(2"a") "f"(x)  "d"x = int_0^"a" "f"(x)  "d"x + int_0^"a" "f"(2"a" - x)  "d"x`