Show that $f(x)=x^n$ is integrable on $[a,b]\quad$where $n\in \mathbb{N}$ And find the value of the integral.
I sincerely cant do the proof i have done when $x^2$ but in this one i have two hints one is that take the partition defined by $x_0=a, x_1=ar, x_2=ar^2,..., x_{n-1}= ar^{n-1}, x_n=ar^n=b$
Where $r= \sqrt[n]{\frac{b}{a}}>1$ (I don't know why is >1 and why that sqrt)
And the other Hint is that makes the difference between upper sums and lower sums and make lower than epsilon to show that is integrable and then choose the infimum of the upper sums and the supremum of the lower sums to find the integral please I'm stuck in this problem. It is not my intention that you make the proof for me but in this one, I'm lost. 
 A: Why don't you follow the hint and calculate the upper und lower sum of these partitions? 
And btw: It's a bad idea to use the same variable for the number of supporting points of your partition then for your function.
To prevent these I will call the partition as:
$x_0=a, x_1=ar, x_2=ar^2,..., x_{m-1}= ar^{m-1}, x_m=ar^m=b$ where $r= \sqrt[m]{\frac{b}{a}}>1$
I will do for the upper sum: 
$$\begin{align*} U(f) &= \sum_{k=0}^{m-1} (x_{k+1} - x_k)\sup_{x \in [x_k,x_{k+1})} f(x) \\ &= \sum_{k=0}^{m-1} (ar^{k+1} - ar^k)x_{k+1}^n \\ &= \sum_{k=0}^{m-1} ar^k(r - 1) \left(ar^{k+1}\right)^n \\ &= a^{n+1}(r-1)r^n \sum_{k=0}^{m-1} \left(r^{n+1}\right) ^k \\\\ &= a^{n+1}(r-1)r^n \frac{\left(r^{n+1}\right)^{m} - 1}{r^{n+1} - 1} \\\\ &= a^{n+1}r^n\frac{\left(r^m\right)^{n+1} - 1}{r^n + r^{n-1} + \ldots + r^0} \\\\ &= a^{n+1}r^n\frac{\left(\frac{b}{a}\right)^{n+1} - 1}{r^n + r^{n-1} + \ldots + r^0} \end{align*}$$
Now consider that the upper sum is monotone in $m$ and hence $$\inf_{m \in N} U(f) = \lim_{m\to\infty} U(f)$$ and that $$\lim_{m\to \infty} r = 1$$ to get:
$$\inf_{m \in N} U(f) = \lim_{m\to\infty} U(f) = a^{n+1}\frac{\left(\frac{b}{a}\right)^{n+1} - 1}{n+1} = \frac{1}{n+1}\left(b^{n+1} - a^{n+1}\right)$$
Do the same for the lower sum to see $$\lim_{m\to\infty} L(f) = \lim_{m\to\infty} U(f)$$ to see the integral exists and then the value is …
