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Math Help - (very simple integral question) WHY, is this answer POS/adding --- PLEASE HELP ---??

  1. #1
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    (very simple integral question) WHY, is this answer POS/adding --- PLEASE HELP ---??

    The problem is: the integral from 1 to e of (x + 1)/(x)

    so you have to distribute the x^-1 to the numerator and you get

    x + 1/x and then you have to integrate

    x/2 + lnx

    and then you do f(b) - f(a)

    (e/2 + 0) - (1/2 + 0) so then its e/2 - 1/2 so isn't the answer (e-1)/2 ??? The online answer says (e+1)/2

    this site (Answers and solutions to 2008 Multiple Choice BC 9-14) says otherwise, claiming that its plus and not minus. *******ITS NUMBER 13**********

    where did I go wrong?!?
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    Re: (very simple integral question) WHY, is this answer POS/adding --- PLEASE HELP --

    It is not "-", it is "+" at (e/2 + 0) - (1/2 + 0)
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    Re: (very simple integral question) WHY, is this answer POS/adding --- PLEASE HELP --

    Quote Originally Posted by ManosG View Post
    It is not "-", it is "+" at (e/2 + 0) - (1/2 + 0)
    Sorry but I don't get what you're saying.

    Why wouldn't it be "-", because when integrating you do the upper limit - the lower limit, after plugging in.

    So since they both have a denominator of 2, it goes on bottom and you subtract the e term and .5 on top.

    I don't get how it could be anything else. Sorry if I'm being hard-headed.
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    Re: (very simple integral question) WHY, is this answer POS/adding --- PLEASE HELP --

    Yes, you are right, my fault, the mistake is \log{e} = 1 \quad not \quad 0
    Thanks from skinsdomination09
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    Re: (very simple integral question) WHY, is this answer POS/adding --- PLEASE HELP --

    Quote Originally Posted by ManosG View Post
    Yes, you are right, my fault, the mistake is \log{e} = 1 \quad not \quad 0
    wowwwww soooooooooooooo stupid. Idiotic.

    THANK YOU SOOOO MUCH. That straight pisses me off lol
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