Recurrence Formula Question - Integration (1 Viewer)

[ratcher0071]

I_n = ∫¹₀ (1- sqrt x)ⁿ dx , n≥0

show I_n = {n/ (n+2)} I_n-1 , n≥1

 

[lolokay]

just use integration by parts (keeping in mind the limits) and then get it in terms of In and In-1

 

[azureus88]

lol thats not much of a clue.

Better clue: (sqrtx) = (sqrtx) -1 + 1 helps to get it in terms of In and In-1

 

[shaon0]

I_n = ∫¹₀ (1- sqrt x)ⁿ dx , n≥0

show I_n = {n/ (n+2)} I_n-1 , n≥1

∫ [0 to 1] (1- sqrt x)n dx , n≥0
= [x(1-sqrt(x))^n]{ 0 to 1} - ∫ [ 0 to 1] x .n(1-sqrt(x))^(n-1).(1/2sqrt(x)) dx
= 0 - (n/2) ∫ [0 to 1] sqrt(x) (1-sqrt(x))^(n-1) dx
= (n/2) ∫ [ 0 to 1] ((1-sqrt(x))-1)(1-sqrt(x))^(n-1) dx
= (n/2) [∫ [ 0 to 1] (1-sqrt(x))^(n) dx - ∫ [ 0 to 1] (1-sqrt(x))^(n-1) dx]
= (n/2) [I{n}-I{n-1}]
2I{n}-nI{n}=-n*I{n-1}
I{n}[n-2]=n*I{n-1}
I{n}=[I{n-1}][n/n-2]