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
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