Thread: Integration, Proof and limit.

$\displaystyle \int_{k}^{k+1} \frac{1}{x} dx, k\geq 0$

Show that $\displaystyle \frac{1}{k+1}\leq ln(\frac{k+1}{k}) \leq \frac{1}{k}$

I've found the integral, but how do i prove?

thank you.

Of the three terms in the inequality, which one corresponds to the exact value of the integral?

middle one.

Right. Is the integrand increasing or decreasing as x increases?

Draw the graph of y= 1/x between x= k and x= k+1. Look at the area of the rectangle with base from k to k+1 and height 1/k, the area of the rectangle with base from k to k+ 1 and height 1/(k+1), and the area under the graph.

By the mean value theorem, there exists a D in [K,K+1] such that $\displaystyle 1/D=(ln(k+1)-ln(k))/(k+1-k)=ln((k+1)/k)$

k_<D<_k+1 so 1/(k+1)_<1/D<_1/k so the required result follows.

Originally Posted by BabyMilo

$\displaystyle \int_{k}^{k+1} \frac{1}{x} dx, k\geq 0$
Show that $\displaystyle \frac{1}{k+1}\leq ln(\frac{k+1}{k}) \leq \frac{1}{k}$

The following is found in most calculus textbooks in a list of properties of integrals.
$\displaystyle \text{If }m \leqslant f(x) \leqslant M\text{ for all }x\in[a,b]\text{ then}$
$\displaystyle m(b - a) \leqslant \int_a^b {f(x)dx} \leqslant M(b - a)$.

Now notice that $\displaystyle K \leqslant x \leqslant K + 1\; \Rightarrow \;\frac{1}{{K + 1}} \leqslant \frac{1}{x} \leqslant \frac{1}{K}.$