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    Hello everyone,
    I'm confused with this question about area under graph.

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    So ok, I integrated correctly and managed to find the area under the graph that is above the X axis, but I'm confused with how to find the shaded area that is below the X axis.
    But we can find the y coordinate of intersection. But we can't consider the shaded area under the X axis as a triangle right?
    Some help is appreciated
    Thanks in advance


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    Integrate the area under the x-axis as normal.. So integrate between 2 and 1 and add your result to the area above the x-axis.
    The difference is that you will get a negative value, which you'll have to change to positive (you can't have negative area)
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    (Original post by fawzi963)
    Hello everyone,
    I'm confused with this question about area under graph.

    Name:  ImageUploadedByTapatalk1397640115.349786.jpg
Views: 97
Size:  84.2 KB

    So ok, I integrated correctly and managed to find the area under the graph that is above the X axis, but I'm confused with how to find the shaded area that is below the X axis.
    But we can find the y coordinate of intersection. But we can't consider the shaded area under the X axis as a triangle right?
    Some help is appreciated
    Thanks in advance


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    No, you just integrate and evaluate from x=_____ to x=________ and take the absolute value (or modulus). Your answer will come out to be negative. Just take the numeric value and add it to the first area (That above the axis).
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    (Original post by Basmaa)
    Integrate the area under the x-axis as normal.. So integrate between 2 and 1 and add your result to the area above the x-axis.
    The difference is that you will get a negative value, which you'll have to change to positive (you can't have negative area)
    Thanks for your reply initially,
    Simple as that?! I did more difficult exercises like those but for some reason I didn't realize this one.

    But if we integrate from 2 to 1 how do we know that it's not the whole area under the graph from 2 to 1 and only the shaded area?


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    (Original post by fawzi963)
    Thanks for your reply initially,
    Simple as that?! I did more difficult exercises like those but for some reason I didn't realize this one.

    But if we integrate from 2 to 1 how do we know that it's not the whole area under the graph from 2 to 1 and only the shaded area?


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    No problemm. Because you're integrating with respect to the equation..

    When you think about it, when you're integrating above the x-axis, how do you know you're only integrating the shaded area?

    You could check out examsolutions.net if you need more help; check the C2 section. The guy explains everything and goes through every exam question step-by-step. Really helped me with mechanics!
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    (Original post by Basmaa)
    No problemm. Because you're integrating with respect to the equation..

    When you think about it, when you're integrating above the x-axis, how do you know you're only integrating the shaded area?

    You could check out examsolutions.net if you need more help; check the C2 section. The guy explains everything and goes through every exam question step-by-step. Really helped me with mechanics!
    Makes sense that we're integrating in respect to the equation, true. To answer your question, what makes me know that I'm finding the area above the X axis because the area under the graph is bounded by the graph it self, not sure if I explained what I meant clearly though.


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    (Original post by m4ths/maths247)
    No, you just integrate and evaluate from x=_____ to x=________ and take the absolute value (or modulus). Your answer will come out to be negative. Just take the numeric value and add it to the first area (That above the axis).
    Thanks for your help sir.


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    (Original post by fawzi963)
    Makes sense that we're integrating in respect to the equation, true. To answer your question, what makes me know that I'm finding the area above the X axis because the area under the graph is bounded by the graph it self, not sure if I explained what I meant clearly though.


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    Lol, my question was part of my explanation.. not really asking a question
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    (Original post by Basmaa)
    Lol, my question was part of my explanation.. not really asking a question
    Haha yea, I realized that both below and above the X axis are bounded by a curve, so what I said initially was makes sense now. Thanks though


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