Question

Prove that `int_0^"a" "f(x)" "dx" = int_0^"a" "f"("a"-"x")"dx"` ,and hence evaluate `int_0^1 "x"^2(1 - "x")^"n""dx"`.

Answer 1

Prove that: `int_0^"a" "f(x)" "dx" = int_0^"a" "f"("a"-"x")"dx"`

Proof: Let
t = a - x
⇒ dt = -dx
When x  = 0, t = a
When x = a, t = 0

Putting the value of x in LHS
∫a0fa-t-dt=-∫a0fa-tdt=∫0afa-tdt=∫0afa-xdx          ∵∫abftdt=∫abfxdx=RHS">

`int_"a"^0 "f"("a"-"t")(-"dt")`

`=- int_"a"^0 "f"("a"-"t")(-"dt")`

`=int_0^"a""f"("a"-"t")(-"dt")`

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

= RHS 

Using this we can solve the given question as follows:

`int_0^1"x"^2(1-"x")^"n" "dx"`

` = int_0^1 (1-"x")^2 (1-(1-"x"))^"n" "dx"`

` = int_0^1 (1+"x"^2 -2"x")("x")^"n" "dx"`

` = int_0^1 ("x"^"n" + "x"^(2+"n")-2"x"^("n"+1))"dx"`

`=["x"^("n"+1)/("n"+1)+ "x"^("n"+3)/("n"+3) - 2 "x"^("n"+2)/("n"+2)]_0^1`

`= [1/("n"+1) + 1/("n"+3) - 2/("n"+2)]`