# Hpw calculate in two ways integral (1-x)^n, if x=0--->1 for show 1 - 1/2nC1 +1/3nC2+...=1/n+1?

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### 1 Answer

You should use binomial theorem to expand `(1-x)^n` such that:

`(1-x)^n = C_n^0*1^n - C_n^1*1^(n-1)*x +....+C_n^n*(-x)^n`

Hence, integrating, yields:

`int_0^1 (1-x)^n dx = int_0^1 C_n^0*1^n dx - int_0^1 C_n^1*1^(n-1)*x dx + ..... + int_0^1 C_n^n*(-x)^n dx`

`int_0^1 (1-x)^n dx = (x - (1/2)C_n^1*x^2 + (1/3)C_n^2*x^2 + .... + (-1)^n*(1/(n+1))C_n^n*x^(n+1))|_0^1`

`int_0^1 (1-x)^n dx = 1 - (1/2)C_n^1 + (1/3)C_n^2 +..+(-1)^n*(1/(n+1))C_n^n`

You need to evaluate `int_0^1 (1-x)^n dx ` changing the variable such that:

`1 - x = y =gt -dx = dy`

`x = 0 =gt y = 1`

`x = 1 =gt y = 0`

`-int_0^1 y^n dy = (-y^(n+1))/(n+1)|_0^1`

`int_0^1 y^n dy = -1/(n+1)`

**Hence, evaluating the integral using both methods yields that `1 - (1/2)C_n^1 + (1/3)C_n^2 +..+(-1)^n*(1/(n+1))C_n^n = -1/(n+1).` **

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