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$\mathbf{EDIT}$: the original definition of the problem had an error. $F(e)$ has to be equal $\frac{5}{3}$ not $1$.

I need to find the antiderivative $F$ of

$f(x) = \begin{cases} x^{-1} \qquad x \ge 1 \\ x^{-2} \qquad 0

such that $F(e) = 1$.

$\mathbf{First:}$

$F(x) = \begin{cases} \ln|x| + c_1 \qquad x \ge 1 \\ -\frac{1}{x} + c_2 \qquad 0

$\mathbf{Secondly}$, we need to make sure that $F(x)$ is continuous at $x=1$:

$$\lim_{x \to 1^+} F(x) = \lim_{x \to 1^-} F(x)$$ Which is: $$\ln1+c_1 = c_2-1 \Rightarrow c_1 = c_2 - 1$$ Therefore:

$F(x) = \begin{cases} \ln|x| + c_2 - 1 \qquad x \ge 1 \\ -\frac{1}{x} + c_2 \qquad 0

$\mathbf{Lastly}$, we need to make sure that $F(e) = 1$:

$$\ln e+ c_2 - 1 = 1$$ $$1+ c_2 - 1 = 1$$ $$c_2 = 1$$

$\mathbf{Finally}$, this is the antiderivative:

$F(x) = \begin{cases} \ln|x| \qquad x \ge 1 \\ -\frac{1}{x} + 1 \qquad 0

I think something is wrong in my calculations because I need to calculate

$F(e^2)+F(e^{-1})+F(e)+F'(0.5)+F'(2)$ where I get $2 -e +1 + 1 + 4 + 0.5 = 8.5-e$ however the answer to the sum is different. The answer can be one of the following: 1) $10.5-e$ 2) $9.5 + e$ 3) $9-e$ 4) $8.5 + 2e$

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    You dont need the absolute value in the logarithm because $x\ge 1$. All the other things seems ok. – 2017-01-19
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    I added some additional details to where my problem is, I thought that would be an issue with my immediate calculation of the antiderivative. – 2017-01-19
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    your calculation is true. – 2017-01-19
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    Do you mean the end sum? – 2017-01-19
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    Thanks, I added the possible answers that I need to choose from. I'm really confused because none of the answers that can be correct don't include mine. – 2017-01-19
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    Yeah it's correct – 2017-01-19
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    I just found out that there was a correction to the exercise and $F(e)=\frac{5}{3}$. Sorry! In this case $10.5 - e$ is correct. – 2017-01-19

1 Answers 1

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You can use the fact that the function $$ G(x)=\int_{1}^x f(t)\,dt $$ is an antiderivative of $f$ on the interval $(0,\infty)$. Then, if $01$, $$ G(x)=\int_{1}^x\frac{1}{t}\,dt=\Bigl[\ln t\Bigr]_1^x=\ln x $$ so our function is $$ G(x)=\begin{cases} 1-\dfrac{1}{x} & \text{if $01$} \end{cases} $$ All other antiderivatives differ from this one by a constant. So you need to find $c$ so that $F(x)=G(x)+c$ has $F(e)=1$; then $$ F(e)=G(e)+c=\ln e+c=1+c $$ and $c=0$.

If you want $F(e)=\frac{5}{3}$, then $$ \frac{5}{3}=G(e)+c=1+c $$ and $c=\frac{2}{3}$.

For this last one, $$ F(e^2)+F(e^{-1})+F(e)+F'(1/2)+F'(2)= 2+\frac{2}{3}+(1-e)+\frac{2}{3}+1+\frac{2}{3}+4+\frac{1}{2}=\frac{21}{2}-e $$