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Math Help - Integration question - 2

  1. #1
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    Integration question - 2



    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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  2. #2
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    Quote Originally Posted by champrock View Post


    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
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  3. #3
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    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?
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