How do you evaluate the definite integral #int dx/x# from #[1/e,e]#?

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Answer 1 · 2016-10-28T12:58:42

2

#int_(1/e)^e(dx)/x=[lnx]_(1/e)^e#

#=lne-ln(1/e)#

#=lne-[ln1-lne]#

#=1-0+1#

#=2#

Answer 2 · 2016-10-28T13:00:34

The answer is #2#

The integral is #intdx/x=lnx + C#

So the integral is #intdx/x=(lnx)#

So from #(1/e,e)#,
we have #intdx/x=lne-ln(1/e)=lne-ln1+lne#

As #lne=1# and #ln1=0#
So we have finally #intdx/x=1-0+1=2#