Integral $\int_1^2 x^{-2}\,dx$ with Darboux sums.

Question

I'm trying to calculate $\int_1^2\frac{1}{x^2}\,dx$, with the partition $2^{\frac{i}{n}}$ , $0\leq i\leq n$. ( Darboux sums)

I tried to solve it for hours but I'm getting strange results.

thanks for help!

Answer 1

The function is continuous and strictly decreasing, so $$U_n=\sum_{k=0}^{n-1}\frac{2^{(k+1)/n}-2^{k/n}}{2^{2k/n}}\\ L_n=\sum_{h=0}^{n-1}\frac{2^{(h+1)/n}-2^{h/n}}{2^{2(h+1)/n}}$$

Thus $$U_n=\sum_{k=0}^{n-1}\frac{2^{1/n}-1}{2^{k/n}}=(2^{1/n}-1)\cdot\frac{2^{-n/n}-1}{2^{-1/n}-1}=(2^{1/n}-1)\frac{-\frac12}{2^{-1/n}(1-2^{1/n})}=2^{(1-n)/n}$$ And, to cut it short, $L_n=2^{-2/n}U_n$