# Integration question - 2

• Mar 10th 2009, 11:20 AM
champrock
Integration question - 2
http://img50.imageshack.us/img50/900...0941pmaaal.png

i think this should be solvable by "by parts" but i am not getting an answer from that. Probably I am not doing it correctly.

thanks
• Mar 10th 2009, 11:41 AM
HallsofIvy
Quote:

Originally Posted by champrock
http://img50.imageshack.us/img50/900...0941pmaaal.png

i think this should be solvable by "by parts" but i am not getting an answer from that. Probably I am not doing it correctly.

thanks

This doesn't ask you to find the integral! ln(x) is an increasing function so for x between 2 and 3, $ln(2)\le ln(x)\le ln(3)$ and $1/ln(3)\le 1/ln(x)\le 1/ln(2)$.

So $1/ln(3)= \int_2^3 dt/ln(3)\le \int_2^3 \frac{dx}{ln(x)}\le 1/ln(2)= \int_2^3 dt/ln(2)$
Now, 1/ln(2)= 1.44 and 1/ln(3)= 0.9102 so that
$0.9102\le \int_2^3 \frac{dx}{ln(x)}\le =1.44$
• Mar 10th 2009, 11:42 PM
champrock
hmm the problem is that i dont have to use a calculator. so i dont know the values of 1/ln 2 and 1/ln 3 .

any completely "manual" way to do this?