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Definite Integrals

  1. Jun 24, 2007 #1

    please help me....

  2. jcsd
  3. Jun 24, 2007 #2


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    You need to show your work before you get help. What have you done with this question?
  4. Jun 24, 2007 #3
    ok well i posted it in the calculus forums but no one helped....
    my friend is making a new website and he told me the only way i could get a peak is by solving this equation, and the reason he did this is becasue he knows far to well i cant do math, I recenty graduated HS and know nothing of calculus, thats why im askign here, if someone would be willing to help me or to even give me crash corse because they dont think its right to just give me the answer i would be more then willing to try and learn.....
  5. Jun 24, 2007 #4
    lol; most odd thing i have ever heard ...
    but, if your story is true, i shall try and help.

    I take it you would know that a square has equal length sides?
    hence the area of the square is equal to the length by width (or in this case [tex] l^2 [/tex]:
    [tex]A(square) = l^2[/tex]
    (if u really need: http://en.wikipedia.org/wiki/Square_(geometry))

    In regard to the next section, i doubt it will be even the slightest bit understandable, but none-the-less ....
    A definite integral basically finds the area underneath a graph in a set region i.e. between a and b: http://en.wikipedia.org/wiki/Image:Integral_as_region_under_curve.svg
    where f(x) is your function, such that: [tex]f(x) = 0.43890022x[/tex]
    but for our purposes, we just want to find the numerical answer for this integral as a means of determining the area of the square.

    The definite integral notation for this is:
    [tex] \int_{1}^{10} {0.43890022x} [/tex]
    This ican be explained as; the definite integral between 1 and 10 for the equation 0.43890022 as required.

    Now we can simplify this integral and obtain a numerical answer:
    [tex] \int_{1}^{10} {0.43890022x} [/tex]
    [tex] = 0.43890022 \int_{1}^{10} {x} [/tex] (taken out factor)
    [tex] = 0.43890022(\frac {x^2}{2}) [/tex] between 1 and 10
    [tex] = 0.43890022 ((\frac {10^2}{2}) - (\frac {1^2}{2})) [/tex]
    plug that into your calculator and you get:
    [tex] 21.72556089 [/tex]

    remember from before:
    [tex]A(square) = l^2[/tex]
    [tex]A(square) = 21.72556089^2[/tex]

    wow that took longer than I thought, but yeh, hope this helps your cause.
    You must understand that you cannot learn how to do definite integration with no understanding of basic calculus methods etc; for the parts you don't understand, just take them as facts ...
    Last edited: Jun 24, 2007
  6. Jun 24, 2007 #5
    Assuming of course that the variable of integration is X and that X varies from 1 to 10 .....
  7. Jun 24, 2007 #6
    hehe, yer ....

    in respect to the problem given; could it be anything else??
    or is that just complicating things more?

  8. Jun 24, 2007 #7
    lol, nah it had to be X, I was just pointing out ways for the OP to get back at his supercilious friend :biggrin:
  9. Jun 24, 2007 #8
    haha, yeh good point and idea ...
    pretty nasty form of entrance don't you think?
  10. Jun 24, 2007 #9

    matt grime

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    As long as you know what 'integral means' calculus is completely irrelevant to the question. It's just a straight line graph - you can find the area by elementary means.
  11. Jun 24, 2007 #10
    omg if that trully is the answer, i got it right when i used one fo the calculators. but double guessed and threw it out since i had no clue. lol wow steven10137 you are my hero.... lol now i have to see what he says, ill come back and post his reply :) thanks a million guys
  12. Jun 25, 2007 #11
    not at all
  13. Jun 25, 2007 #12
    lol its awsome ty again, he was shocked but gave in after i got angry LOL, thank you all again
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