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Evaluate definite integral. x if x<1; 1/x if x> or equal to 1.

  1. Jan 12, 2013 #1
    Evaluate definite integral. "x if x<1; 1/x if x> or equal to 1."

    1. Consider the function: f(x) = {x if x<1
    {1/x if x≥1

    Evaluate the definite integral.

    ∫from 0 to 4 of f(x)dx


    2. Okay, I think I vaguely remember something about these sorts of problems... Isn't it something like you choose which value to go along with depending on what the limits of the integral are?
    But the limit 0 goes along with "x if x<1" and 1 through 4 go along with "1/x if x≥1."
    And integrals just evaluate area, so can we break this up into 2 integral problems?

    Integral #1

    ∫0 to 0
    Wait...But that would be just zero. If the limits were from 0 to 0, then there'd be no area! :/

    Integral #2

    ∫1 to 4 of (1/x)dx

    = ln(4) - ln(1) = 1.386

    Yeah, this is not the right answer.
    I don't even really know what I'm doing... =_=
    I'm probably not even going about it right at all...
    Help?
    Thank you SO much!
     
  2. jcsd
  3. Jan 12, 2013 #2

    haruspex

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    Re: Evaluate definite integral. "x if x<1; 1/x if x> or equal to 1."

    Right approach, but where did you get "0 to 0" from in the first integral?
     
  4. Jan 13, 2013 #3
    Re: Evaluate definite integral. "x if x<1; 1/x if x> or equal to 1."

    I was thinking that the "x if x<1" meant that 1 could not be part of the limit...

    But maybe it's okay to have the integral from zero to one because it's just the area right up to one?

    So would it be instead:

    Integral #1

    ∫0 to 1 of x

    = (x^2)/2 |0 to 1

    = (1^2)/2 - (0^2)/2

    = 1/2 - 0 = 1/2

    Integral #2

    ∫1 to 4 of (1/x)dx

    = ln(4) - ln(1) = 1.386

    So would the answer be .5 + 1.386 = 1.886 ?
    Not sure at all... :/
     
  5. Jan 13, 2013 #4

    Ray Vickson

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    Re: Evaluate definite integral. "x if x<1; 1/x if x> or equal to 1."

    The inequality x < 1 just means that '1' is excluded, but everything to the left of 1 is included. The area under the curve AT x = 1 is zero (that is, it is the area of a line segment, which = 0). So, yes: the first part is the integral from 0 to 1.
     
  6. Jan 13, 2013 #5

    SammyS

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    Re: Evaluate definite integral. "x if x<1; 1/x if x> or equal to 1."

    If you want to get picky, for Integral#1 you could do

    [itex]\displaystyle \lim_{\,a\to1^-\,}\int_0^a x\,dx\ .[/itex]

    But that will give the same result as [itex]\displaystyle \ \ \int_0^1 x\,dx\ .[/itex]
     
  7. Jan 13, 2013 #6

    HallsofIvy

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    Re: Evaluate definite integral. "x if x<1; 1/x if x> or equal to 1."

    Strictly speaking, as SammyS said, you can't. You need to use the limit as an improper integral. But even so, surely you realize that there are numbers between 0 and 1?

     
  8. Jan 13, 2013 #7
    Re: Evaluate definite integral. "x if x<1; 1/x if x> or equal to 1."

    Okay, so it's okay to have the limits from 0 to 1 and from 1 to 4.

    But my hw computer program does not count the answer 1.886 as correct...

    Did I make some mistake in calculating the integrals?
    Do you have any idea where I went wrong?

    Thank you!
     
  9. Jan 13, 2013 #8

    SammyS

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    Re: Evaluate definite integral. "x if x<1; 1/x if x> or equal to 1."

    Maybe it wants the exact number, i.e. (1/2)+ln(4) .
     
  10. Jan 13, 2013 #9
    Re: Evaluate definite integral. "x if x<1; 1/x if x> or equal to 1."

    @SammyS - Wow! You were exactly right! Thank you so much! :D
     
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