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Math Help - [SOLVED] Online Definite Integral Calculators faulty?

  1. #1
    dtb
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    [SOLVED] Online Definite Integral Calculators faulty?

    Hi all,

    I've been looking at two online calculators of Definite Integrals and I'm wondering if they can't handle limits that cross the x-axis

    eg.
    \int_{-3}^{1} \! sin(x) \, dx.

    My solution is that the area is 2.4486 (being 1.989 + 0.459 area below axis added to area above axis)

    But if you use either of these two popular online Integral calculators
    Find the Numerical Answer to a Definite Integral - WebMath
    Definite Integral Calculator at SolveMyMath.com

    they give a result of -1.53 (negative area subtract smaller positive area)

    So are they wrong? or is there a different purpose for this negative result?

    Cheers
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by dtb View Post
    Hi all,

    I've been looking at two online calculators of Definite Integrals and I'm wondering if they can't handle limits that cross the x-axis

    eg.
    \int_{-3}^{1} \! sin(x) \, dx.

    My solution is that the area is 2.4486 (being 1.989 + 0.459 area below axis added to area above axis)

    But if you use either of these two popular online Integral calculators
    Find the Numerical Answer to a Definite Integral - WebMath
    Definite Integral Calculator at SolveMyMath.com

    they give a result of -1.53 (negative area subtract smaller positive area)

    So are they wrong? or is there a different purpose for this negative result?

    Cheers
    When calculating an area, you have to subtract any area that is below the x-axis. And add the areas that are above the x-axis.
    That's integrals, they can be negative, though areas in geometry are always positive.


    If you work with the antiderivatives... :
    An antiderivative of sin(x) is -cos(x) (just check by taking the derivative of -cos(x))
    By definition, your integral is thus -\cos(1)-[-\cos(-3)]=-\cos(1)+\cos(-3)=-\cos(1)+\cos(3)
    You can check this result with a calculator, it'll give -1.53
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  3. #3
    dtb
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    Hmm, I'm still not sure...

    I can work it out that way if I leave all +/- symbols as they are:

    I get
    -1.9899 (area below x-axis) + 0.4597 (area above x-axis)
    = -1.5302

    Unfortunately, having looked at 5 calculus books, there isn't much mention of graphs that cross the x-axis....

    My teacher was saying that you need to take each area separately - make a negative area positive - then add them together
    ie. magnitude |-1.9899| + |0.4597|
    = 1.9899 + 0.4597
    = 2.4496

    So the area is positive, and a sum of the part below the axis and the part above.

    Any thoughts?
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  4. #4
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    If f is nonnegative and integrable on [a,b] then \int_a^b {f(x)dx} is a measure of the area bound by the graph of f and the x-axis.

    Therefore, \int_0^{\frac{\pi }{2}} {\cos (x)dx}  = 1 means that there is 1 square unit in the region bound by \cos(x) and the x-axis from x=0 to x=\frac{\pi}{2}.

    However, \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\cos (x)dx}  = 0 this is not area because the function is not nonnegative on that entire interval.

    But \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\left| {\cos (x)} \right|dx}  = 2 which is the correct bounded area.
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  5. #5
    dtb
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    I figured out why I was getting confused...

    There's a difference between the "integral" and "the area between the graph and the x-axis"

    or at least there is where the graph crosses the x-axis.

    So, those online calculators will evaluate an integral, but they won't tell you the area.

    Now I just need to find out what each type of result is used for
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  6. #6
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    Quote Originally Posted by dtb View Post
    There's a difference between the "integral" and "the area between the graph and the x-axis"
    Now I just need to find out what each type of result is used for
    Read my reply. The use if absolute value will give area.
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  7. #7
    dtb
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    Nah I was just wondering when/why in real world situations one might one evaluate the integral vs. find the area, that's all
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