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Determine the truth of the following statements

  1. Mar 12, 2017 #1

    diredragon

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    1. The problem statement, all variables and given/known data
    ##f(x) =
    \begin{cases}
    -\frac{1}{1+x^2}, & x \in (-\infty,1) \\
    x, & x \in [1,5]\setminus {3} \\
    100, & x=3 \\
    \log_{1/2} {(x-5)} , & x \in (5, +\infty)
    \end{cases}##
    For a given function determine the truth of the folowing statements and give a brief explanation:
    a) Function ##F(x)=\arctan {1/x}## is one integral solution of the funtion ##f(x)## on an interval of ##(-1,1)##
    b) Function f is integrable on ##[-3,5]##
    c) Function f is integrable on ##[4,10]##
    d) ##\int_{-1}^{1} \tan x f(x) \, dx=0##
    2. Relevant equations
    3. The attempt at a solution

    a) Function ##F(x)=\arctan {1/x}## is one integral solution of the funtion ##f(x)## on an interval of ##(-1,1)##
    False. The solution of this integral on a provided interval is ##F(x)=-\arctan x + c##
    b) Function f is integrable on ##[-3,5]##
    Yes. It has 3 points of break and is continuous on intervals in between so yes. Not sure about this one so i would be grateful for your reply
    c) Function f is integrable on ##[4,10]##
    Yes. The same reasoning as above.
    d) ##\int_{-1}^{1} \tan x f(x) \, dx=0##
    Yes, the function is odd.
     
  2. jcsd
  3. Mar 13, 2017 #2

    haruspex

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    For a), it would be safer to consider the derivative of F. Integration can lead to multiple results that look different but are all valid solutions.

    For c), consider what the function does as it tends to 5 from above.

    I agree on d).
     
  4. Mar 13, 2017 #3

    diredragon

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    Hmm.. It seems that the possible solutions to the first one involve three functions that look different but differ only by a constant. ##-\arctan x+c##, ##\cot^{-1} x+c## and ##-\arctan(1/x)+c## so it is tru then.
    Ooh, so for the part c) you wanted me to consider what happens to the function as x approaches 5 and i should have spotted that on the left as x approaches 5 and goes near it, my function goes from ##-\infty##. So the problem part c) has that part b) doesnt have is that the function on its points of break can be integrated as it doesnt go to infinity on those points? Did i get this right?
     
  5. Mar 13, 2017 #4

    haruspex

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    Right. Of course, it might still be integrable, like ∫011/√x, so there is more work to do.
    Yes, so the justification you gave for b) was incomplete.
     
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