Integration, Proof and limit.

**BabyMilo** · Jun 15th 2011, 01:23 AM

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

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

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

thank you.

**Ackbeet** · Jun 15th 2011, 01:32 AM

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

**BabyMilo** · Jun 15th 2011, 01:34 AM

middle one.

**Ackbeet** · Jun 15th 2011, 01:35 AM

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

**HallsofIvy** · Jun 15th 2011, 01:40 AM

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.

**boromir** · Jun 15th 2011, 01:46 AM

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

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

**Plato** · Jun 15th 2011, 08:41 AM

Originally Posted by BabyMilo

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

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

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

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