Question

Prove that: `int_0^"a" "f"(x)  "d"x = int_0^"a" "f"("a" - x)  "d"x`. Hence find `int_0^(pi/2) sin^2x  "d"x` 

Answer 1

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

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

Put a – x = t

∴ – dx = dt

∴ – dx = dt

When x = 0, t = a – 0 = a

and when x = a, t = a – a = 0

∴ I = `int_4^0 "f"("t")(-"dt")`

= `-int_"a"^0 "f"("t")  "dt"`

= `int_0^"a" "f"("t")  "dt"`    .......`[∵ int_"a"^"b" "f"(x)  "d"x = -int_"b"^"a" "f"(x)  "d"x]`

= `int_0^"a" "f"(x)  "d"x`   .......`[∵ int_"a"^"b" "f"(x)  "d"x = int_"a"^"b"  "f"("t")  "dt"]` 

= L.H.S.

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

Let I = `int_0^(pi/2) sin^2x  "d"x`     .......(i)

= `int_0^(pi/2) sin^2(pi/2 - x)  "d"x`    .......`[∵ int_0^"a" "f"(x)  "d"x = int_0^"a"  "f"("a" - x)  "d"x]`

∴ I = `int_0^(pi/2) cos^2  "d"x`     .......(ii)

Adding (i) and (ii), we get

2I = `int_0^(pi/2) sin^2x  "d"x + int_0^(pi/2) cos^2x  "d"x`

= `int_0^(pi/2) (sin^2x + cos^2x)  "d"x`

∴ 2I = `int_0^(pi/2)1* "d"x`

∴ I = `1/2[x]_0^(pi/2)`

∴ I = `1/2(pi/2 - 0)`

∴ I = `pi/4`