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Range of Integrated function

  1. May 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##g(x) = \int_0^xf(t) dt## where ##f## is such that ##\frac{1}{2} \leq f(t) \leq 1## for ##t \in [0,1]## and ##\frac{1}{2} \geq f(t) \geq 0## for ##t \in (1,2]##. Then ##g(2)## belongs to interval
    A. ##[\frac{-3}{2}, \frac{1}{2}]##
    B. ##[0, 2)##
    C. ##(\frac{3}{2}, \frac{5}{2}]##
    D. ##(2, 4)##

    2. Relevant equations


    3. The attempt at a solution
    I got ##g'(x) = f(x)## and using this and the definite integral given, i have ##g(0) = 0##
    I didn't really know where to go from here, so I tried making a graph (sort of) using the minimum and maximum slopes of the function in the given intervals and found an area in which, I think the function will exist, with the interval for ##g(2)## being ##[\frac{1}{2},\frac{3}{2}]##. ^CEEE41B8BDC6F80CABC7D6937AE5C0255667098491E38C8A1F^pimgpsh_fullsize_distr.jpg
    This isn't present in the options...Can someone please point out my mistakes and help me get the answer.
     
    Last edited: May 4, 2017
  2. jcsd
  3. May 4, 2017 #2

    BvU

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    I mostly agree with your reasoning; generally the lower bound of an interval is given first, followed by the upper bound: ##[{1\over 2},{3\over 2}]## is your result. If it isn't in the list litterally, you might try to exclude the answers that certainly don't satisfy...
     
  4. May 4, 2017 #3
    Ah silly me. Will edit it. I found the answer key too and it says B is correct. Is ##g(0) =0## correct?
     
  5. May 4, 2017 #4

    BvU

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  6. May 4, 2017 #5
    Now I get it! Thank you!
     
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