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प्रश्न
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`
उत्तर
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`
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