$f(x)=2^x$. $x \in [0,10]$ Darboux sums

Question

Dividing given interval onto $n$ equal parts find Darboux upper and lower sums.

$f(x)=2^x$. $x \in [0,10]$

$L(f,P)$ = $\frac{10}{n}\sum_{i=1}^{n}2^\frac{10(i-1)}{n}$

$U(f,P)$ = $\frac{10}{n}\sum_{i=1}^{n}2^\frac{10i}{n}$

I can't continue from here,I don't think I did something wrong.

Answer 1

Although the "right way" is to use the Geometric Series Formula, you actually can argue even without that.

Just note that $$U(f,P)-L(f,P)=\frac{10}{n}\sum_{i=1}^{n}2^\frac{10i}{n}-\frac{10}{n}\sum_{i=1}^{n}2^\frac{10(i-1)}{n}$$ which means $$U(f,P)-L(f,P)=\frac{10}{n}\left(\sum_{i=1}^{n}2^\frac{10i}{n}-\sum_{i=1}^{n}2^\frac{10(i-1)}{n}\right)$$

I'll leave it to you to find out which terms will get cancelled, which ones will remain, and what will happen to the remaining terms when you take $\lim_{n\to \infty}$. If you can't figure it out, do tell me and I'll add it to my answer.

Answer 2

I hope my way is right. We use the formula for the geometric series and the definition of the integral.

First we have to split $I=[a,b]$. We use $x_k=a+\frac{b-a}{n+1}k$. For our problem: $x_k=\frac{10}{n+1}k$. Now we put this $x_k$ into our function $f(x)=2^x \Rightarrow f(x_k)=2^{\frac{10}{n+1}k}$. Using the definition of darboux-sums we find: $$L=\sum_{k=0}^n 2^{\frac{10}{n+1}k}\cdot \frac{10}{n+1}=\frac{10}{n+1} \sum_{k=0}^n 2^{(\frac{10}{n+1})^k}$$ Now we can use the formula for the geometric series: $\sum_{k=0}^n q^k=\frac{1-q^{n+1}}{1-q} $ with $q=2^{\frac{10}{n+1}}$

$$...=\frac{10}{n+1}\left(\frac{1-\left(2^\frac{10}{n+1}\right)^{n+1}}{1-2^\frac{10}{n+1}} \right)=\frac{10}{n+1}\left(\frac{1-2^{10}}{1-2^{\frac{10}{n+1}}} \right):= s_n$$

The last step ist to evaluate the limit when $n \to \infty$. So we finaly get $$\lim_{n\to \infty} s_n=\frac{1023}{\log(2)}\approx 1475,87703 $$