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

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- Mar 10th 2009, 10:20 AMchamprockIntegration 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, 10:41 AMHallsofIvy
This doesn't ask you to find the integral! ln(x) is an increasing function so for x between 2 and 3, $\displaystyle ln(2)\le ln(x)\le ln(3)$ and $\displaystyle 1/ln(3)\le 1/ln(x)\le 1/ln(2)$.

So $\displaystyle 1/ln(3)= \int_2^3 dt/ln(3)\le \int_2^3 \frac{dx}{ln(x)}\le 1/ln(2)= \int_2^3 dt/ln(2)$

Now, 1/ln(2)= 1.44 and 1/ln(3)= 0.9102 so that

$\displaystyle 0.9102\le \int_2^3 \frac{dx}{ln(x)}\le =1.44$ - Mar 10th 2009, 10:42 PMchamprock
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?