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