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# Calculus Question watch

1. 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.

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I get Area=-4/3. The negative sign indicates that the area is below the x-axis.
3. I'm getting 4/3. Integral is (1/3)x^3 - x.
4. I get -4/3. So I just make it positive?
5. (Original post by Unknown?)
I get -4/3. So I just make it positive?
6. 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.
7. whether it is positive or negative depends on whether it is looking for signed or geometric integration
8. (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.
9. (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.
10. 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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Updated: February 4, 2010
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