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

The problem is: the integral from 1 to e of (x² + 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

x²/2 + lnx

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

(e²/2 + 0) - (1/2 + 0) so then its e²/2 - 1/2 so isn't the answer (e²-1)/2 ??? The online answer says (e²+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?!? (Angry)(Angry)(Angry)(Angry)(Angry)(Angry)(Angry)( Angry)

Re: (very simple integral question) WHY, is this answer POS/adding --- PLEASE HELP --

It is not "-", it is "+" at (e²/2 + 0) - (1/2 + 0)

Re: (very simple integral question) WHY, is this answer POS/adding --- PLEASE HELP --

Quote:

Originally Posted by

**ManosG** It is not "-", it is "+" at (e²/2 + 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.

Re: (very simple integral question) WHY, is this answer POS/adding --- PLEASE HELP --

Yes, you are right, my fault, the mistake is $\displaystyle \log{e} = 1 \quad not \quad 0$

Re: (very simple integral question) WHY, is this answer POS/adding --- PLEASE HELP --

Quote:

Originally Posted by

**ManosG** Yes, you are right, my fault, the mistake is $\displaystyle \log{e} = 1 \quad not \quad 0$

wowwwww soooooooooooooo stupid. Idiotic.

THANK YOU SOOOO MUCH. That straight pisses me off lol