Show that:

the integral between 1 and 0 of(1+(a-1)x)^n= (a^(n+1)-1)/{(n+1)(a-1)}

where a isn't 1 and n is a positive integer.

Please help. I have no idea how to intergrate it.

Thanks

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- March 4th 2007, 11:11 PMfree_to_flyDefinite integral
Show that:

the integral between 1 and 0 of(1+(a-1)x)^n= (a^(n+1)-1)/{(n+1)(a-1)}

where a isn't 1 and n is a positive integer.

Please help. I have no idea how to intergrate it.

Thanks - March 4th 2007, 11:32 PMJhevon
First, let's do the integral: (note, you just treat a - 1 as a constant and do it as you would any other integral)

int{(1+(a-1)x)^n}dx

We proceed by substitution:

Let u = 1+(a-1)x

=> du = (a - 1) dx

=> 1/(a - 1)du = dx

so our integral becomes:

1/(a - 1)*int{u^n}du

= 1/(a - 1){u^(n + 1)]/(n + 1)} + C

= 1/(a - 1){[(1 + (a - 1)x)^(n + 1)]/(n + 1)} + C

Now we evaluate our answer between 0 and 1.

1/(a - 1)*{[(1 + (a - 1))^(n + 1)/(n + 1)] - [(1 + (a - 1)(0))^(n + 1)/(n + 1)]}

= 1/(a -1)*{a^(n+1)/(n+1) - 1^(n+1)/(n+1)}

= 1/(a -1)*{a^(n+1)/(n+1) - 1/(n+1)}

= 1/(a - 1)*((a^(n+1) - 1)/(n + 1))

= (a^(n+1)-1)/{(n+1)(a-1)} as desired - March 5th 2007, 02:23 AMfree_to_flyThanks
Thanks so much for your help. Now I've avoided the shrill screams from my maths teachers. Thanks:)