Evaluation of $\displaystyle \int^{2}_{1}\frac{1}{x}dx$ using limit as a sum
Try: Using The formula $$\int^{b}_{a}f(x)dx = \lim_{h\rightarrow 0}h\times \sum^{n-1}_{r=1}f(a+rh)$$
where $nh=b-a$
above $a=1,b=2$ and $\displaystyle f(x)=\frac{1}{x}$ and $nh=1$
$$\int^{2}_{1}\frac{1}{x}dx = \lim_{h\rightarrow 0} \sum^{n-1}_{r=1}f(a+rh)=\lim_{h\rightarrow 0}h\cdot \sum^{n-1}_{r=0}f(1+rh) $$
$$\lim_{h\rightarrow 0}h\cdot \sum^{n-1}_{r=0}\frac{1}{1+rh}=\lim_{h\rightarrow 0}\bigg[\frac{h}{1+h}+\frac{h}{1+2h}+\cdots \cdots +\frac{h}{1+(r-1)h}\bigg]$$
i did not know how i proceed, struck here
could some help me. thanks
Best Answer
Consider the points $x=c^k$ with $c^n=2$.
$$\int_1^2\frac{dx}{x}\approx\sum_{k=0}^{n+1} \frac{\Delta c^k}{c^k}=\sum_{k=1}^n \frac{ c^{k+1}-c^k}{c^k}=n(c-1)=n\left(\sqrt[n]2-1\right).$$
Then,
$$\lim_{n\to\infty}n\left(\sqrt[n]2-1\right)=\lim_{h\to0}\frac{2^h-1}h=\left.(2^h)'\right|_{h=0}=\log2.$$
Note that this is in fact a discrete version of an exponential change of variable, $x=e^t$, giving
$$\int_1^2\frac{dx}x=\int_{\log1}^{\log2}\frac{e^t\,dt}{e^t}=\int_0^{\log2}dt.$$
The latter integral can be trivially computed as a sum.