Question

Prove that: `int_"a"^"b" "f"(x)  "d"x = int_"a"^"b" "f"("a" + "b" - x)  "d"x`

Answer 1

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

Let I = `int_"a"^"b" "f"("a" + "b" - x)  "d"x`

Put a + b – x = t

∴ – dx = dt

∴ dx = – dt

When x = a, t = a + b – a = b

and when x = b, t = a + b – b = a

∴ I = `int_"b"^"a" "f"("t")(-"dt")`

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

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

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

= L.H.S.

∴ `int_"a"^"b" "f"(x)  "d"x = int_"a"^"b" "f"("a" + "b" - x)  d"x`