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This problems are kind of simple but I don't know how to approach them.So I know the following things $$ 1)F'(x)=1/x $$for every real x\0 $$ F(-1)=1$$ $$F(1)=0 $$ $$F(e)+F(-e)=?$$ F'(x) is f(x) so I know that my function is f(x)=1/x so I also know that F(x)=lnx.From $$F(1)=0$$ I find that $$c=0$$ but what about F(-1)? I can't have ln(-1),what am I doing wrong? Another similar problem is $$f(x)=e^{x^2}$$ and $$F(-1)=0$$

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  • $\begingroup$ For the second problem the answers in my textbook are: $$A.F(1)<0 $$$$B.F(1)=0 $$$$C.F(1)>2$$$$D. F(1)=2$$ $\endgroup$ – Lola May 9 '17 at 10:53
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    $\begingroup$ You have already found that $F(X) = \ln(x) + C$ with $C = 0$ for $x>0$. Try to use the same approach for $x<0$. $\endgroup$ – Marc May 9 '17 at 10:55
  • $\begingroup$ Could you be more explicit? $\endgroup$ – Lola May 9 '17 at 10:59
  • $\begingroup$ Hint: Something of the sort $\ln(-x)$. $\endgroup$ – Marc May 9 '17 at 11:01
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    $\begingroup$ Actually, if $f(x)=\frac1x$, then $F(x)=\log|x|+c$ (note it's the absolute value of $x$ rather than just $x$ in the $\log$) $\endgroup$ – vrugtehagel May 9 '17 at 11:08
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As the domain of $\frac{1}{x}$ (i.e. $\mathbb{R}\setminus \{0\}$) has two parts (connected components, technically) this means that the integration constants can be different on both components:

$F(x) = \ln(x) + C_1$ for $x > 0$ and $F(x) = \ln(-x) + C_2$ for $x <0$.

$F(-1) = 1$ implies $\ln(1)+ C_2 = 1 \text{ so } C_2 = 1$. $F(1) = 0$ implies $\ln(1) + C_1 = 0$ so $C_1 =0$.

Now $F(e) + F(-e) = (\ln(e) + C_1) + (\ln(e) + C_2) = 3$.

As to the second problem, all I could think of was: if $F$ is the primitive of $e^{x^2}$ and $F(-1) =0$ we know $$F(x) = F(x) - F(-1) =\int_{-1}^x e^{x^2}dx$$

So $$F(1) = \int_{-1}^{1} e^{x^2} dx = 2\int_0^1 e^{x^2} dx$$

which is at least strictly positive.. and Wolfram alpha tells us it's$> 2$ in fact (maybe using a lower bound function with computable primitive would tell us the same thing)

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  • $\begingroup$ Sorry for asking this just now but from what I understand,related to integrals, if I have two intervals for x as I have above x>0 and x<0 does it mean I have 2 different constants?For example if I have x is in (-1,0) and x is in (5,8) and (100,900),lets say,does that mean I'll have 3 constants or just 2 for positive and negative x?Sorry if my answer is silly but im trying to understand integrals $\endgroup$ – Lola Jul 4 '17 at 6:10

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