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    Is the magnitude of the area between the graph and the x-axis given by the integral with respect to x of the modulus of the equation of the graph? e.g. While the integral of the normal equation would sum the area over the x-axis and subtract the area beneath it, the integral of the modulus of the equation sums the area over the x-axis and adds this to the area beneath it.
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    (Original post by Big-Daddy)
    Is the magnitude of the area between the graph and the x-axis given by the integral with respect to x of the modulus of the equation of the graph? e.g. While the integral of the normal equation would sum the area over the x-axis and subtract the area beneath it, the integral of the modulus of the equation sums the area over the x-axis and adds this to the area beneath it.
    Yes, this is correct.

    Hence, to work out the integral of the modulus of a function, you need to:

    Find all points of intersection with the chosen axis (x), omitting points of inflection.

    Find weather each region lies below or above the chosen axis (x).

    Split the integral into a series of integrals that alternates between positive and negative (taking care to adjust the limits correctly).

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    (Original post by Jkn)
    Yes, this is correct.

    Hence, to work out the integral of the modulus of a function, you need to:

    Find all points of intersection with the chosen axis (x), omitting points of inflection.

    Find weather each region lies below or above the chosen axis (x).

    Split the integral into a series of integrals that alternates between positive and negative (taking care to adjust the limits correctly).

    Thanks, yes, I knew the process, I just wanted to find out if it corresponded directly with the "integral of the modulus" (because my calculator can integrate the modulus directly and as a back-up in case I'm running out of time this is very useful).
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    (Original post by Big-Daddy)
    Thanks, yes, I knew the process, I just wanted to find out if it corresponded directly with the "integral of the modulus" (because my calculator can integrate the modulus directly and as a back-up in case I'm running out of time this is very useful).
    Okay, no worries. It's a very useful thing to know
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    (Original post by Jkn)
    Find weather each region ...
    Apologies for hijacking someone else's thread, but I know you're usually very particular about the use of language (not to mention an A level English student!) and I've spotted this about 4 times in your threads now without saying anything:

    'weather' = sun/rain/snow etc

    I think the word you want is 'whether'
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    (Original post by davros)
    Apologies for hijacking someone else's thread, but I know you're usually very particular about the use of language (not to mention an A level English student!) and I've spotted this about 4 times in your threads now without saying anything:

    'weather' = sun/rain/snow etc

    I think the word you want is 'whether'
    Well that's embarrassing

    Rest assured, I do know that! I think I type too quickly :lol: Thanks! :lol:
 
 
 
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