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Thread: Integration Concern, the total area of x^3

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    Cadet cloud edwards's Avatar
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    Integration Concern, the total area of x^3

    So i'm stuck with this idea in my head and asked a couple of mates what the area is of the x^3 between x+-1 and x=1.

    One of me friends said the area is 0units^2 from following: [x^-4/4] between x=-1 and 1. = 1-1=0.

    Another said it was 2 units^2 as he did: [x^-4/4] between x=-1 and 1 using absolute values, so the total was 1+1=2.

    I understand how both of me friends came to their conclusion, but i'm still not sure particularly with the thought of the cubic graph having an area of 0 seems weird. So which one is right? By using the absolute value and getting 2units^2 the 0 units.

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    Re: Integration Concern, the total area of x^3

    Quote Originally Posted by cloud edwards View Post
    So i'm stuck with this idea in my head and asked a couple of mates what the area is of the x^3 between x+-1 and x=1.

    One of me friends said the area is 0units^2 from following: [x^-4/4] between x=-1 and 1. = 1-1=0.

    Another said it was 2 units^2 as he did: [x^-4/4] between x=-1 and 1 using absolute values, so the total was 1+1=2.

    I understand how both of me friends came to their conclusion, but i'm still not sure particularly with the thought of the cubic graph having an area of 0 seems weird. So which one is right? By using the absolute value and getting 2units^2 the 0 units.
    If you want the actual physical area, it's the one using absolute values.

    The thing that is 0 is the signed area, or net area.

    If we just say "area" of a function on an interval without saying "signed area" or "net area", we would probably be referring to the physical area, which is always positive for a continuous nonzero function.
    Last edited by InteGrand; 23 May 2017 at 12:20 AM.

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    Re: Integration Concern, the total area of x^3

    Following on from what Integrand said, your first friend (0 units) just evaluated the integral, your second friend (2 units) found the area.

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