We have #F:RR#*#->RR# such that #F'(x)=1/x,F(-1)=1,F(1)=0#.How to calculate #F(e)+F(-e)#?

1 Answer
Jim H
Jun 8, 2017

The antiderivative has two parts (pieces, cases).

Explanation:

From the fact that we are given #F(-1)# and #F(1)#, I conclude that the domain includes positive and negative real numbers.

#F'# however has a discontinuity at #0#, so we must allow that the antiderivative may be defined by cases (piecewise) with different constants on each piece.

From #F'(x) = 1/x#, we conclude that

#F(x) = {(lnx+C_1,x > 0),(ln(-x)+C_2,x < 0):}#

Now use #F(-1) = 1# to conclude that #C_2 = 1#

and #F(1) = 0# tells us that #C_1 = 0#.

So we have

#F(x) = {(lnx, x > 0),(ln(-x)+1,x < 0):}#

Finally,

#F(e)+F(-e) = ln(e)+(ln(e)+1) = 3#

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