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    Rather easy question but I just need a reminder on how to do it as I haven't done these in a while.
    I have to find the area enclosed by the curve X^2 -1. The area enclosed is going to be below the x axis so first I intergrated the equation and substituted 1 into the equation as it crosses the x axis at 1 and then doubled it since the curve is symmterical and also crosses through at -1 but I can't seem to get the right answer. Can someone just check this for me please. :o:
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    What's your answer?

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    I get Area=-4/3. The negative sign indicates that the area is below the x-axis.
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    I'm getting 4/3. Integral is (1/3)x^3 - x.
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    I get -4/3. So I just make it positive?
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    (Original post by Unknown?)
    I get -4/3. So I just make it positive?
    The negative sign indicates that the area is below the x-axis.
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    Okay thankyou, what about this following question, its a little more complicated. I have to find the area bounded by x^3 -x^2 -2x and the x-axis.

    I first divided the equation to get the roots 2, 0 and -1 and then intergrated the equation and substitued 2 and -1 into the new equation, made the negative area postive and added them together.

    I got 19/2 +8/3 = 17/4 but its apparently not the correct answer.
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    whether it is positive or negative depends on whether it is looking for signed or geometric integration
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    (Original post by Unknown?)
    Okay thankyou, what about this following question, its a little more complicated. I have to find the area bounded by x^3 -x^2 -2x and the x-axis.

    I first divided the equation to get the roots 2, 0 and -1 and then intergrated the equation and substitued 2 and -1 into the new equation, made the negative area postive and added them together.

    I got 19/2 +8/3 = 17/4 but its apparently not the correct answer.
    8/3 is correct, 19/2 should be 5/12.
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    (Original post by xXx_pUnK_pOp_pRiNcEsS_xXx)
    8/3 is correct, 19/2 should be 5/12.
    Thankyou I got it right now. I keep making mistakes in my working out even though I got the right method.
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    Also I was just wondering I haven't done a question like this yet but if I come across a question where it asks me to find the area between two curves do I find the area's where they intersect first, then intergrate both equations and subsititue the values in and take them away from each.

    For example if two curves intersected at 3 and 5 (I'm just making this up) would I subtract the equations the both curves away from each other. Intergrate the new curve and substitute the value I got and take the value using 3 away from the value using 5?
 
 
 
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