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anonymous

  • one year ago

Find the area of the region bounded by the curves y = x^2 − 1 and y = sin(x). Give your answer correct to 2 decimal places.

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  1. anonymous
    • one year ago
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    not sure what to do about the graph being partially in the negative quadrants

  2. anonymous
    • one year ago
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    i found the endpoints of intersection at -0.636733 and 1.40962 not sure where to go on from there

  3. anonymous
    • one year ago
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    @ganeshie8 @dan815 @Michele_Laino @pooja195 @Kainui

  4. anonymous
    • one year ago
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    do i have to integrate with respect to the y-axis ?

  5. IrishBoy123
    • one year ago
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    hope this helps you

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  6. anonymous
    • one year ago
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    i had already graphed it :(

  7. IrishBoy123
    • one year ago
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    cool found where they intersect?

  8. anonymous
    • one year ago
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    yep at x = -0.636733 and 1.40962

  9. anonymous
    • one year ago
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    not sure what to do next ...

  10. IrishBoy123
    • one year ago
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    well, next integrate :p i'd be sure to keep the +ve and -ve signs in order so you might wish to do 2 integrations. they cross at x = 0. so integrate x = -0.636733 to 0, and x = 0 to 1.40962 all in dx you can do dy if you want but you are creating a whole load of problems for yourself such as having to integrate arcsin

  11. anonymous
    • one year ago
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    wouldn't the fact that some of the graph is on the negative quadrants give me a wrong answer

  12. IrishBoy123
    • one year ago
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    yep, that's exactly what you have to look out for i'll try draw something

  13. IrishBoy123
    • one year ago
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    |dw:1438890387798:dw|

  14. IrishBoy123
    • one year ago
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    it's a real drag but you will need a way to keep the sign consistent, the areas under the curve will come out as negative. i think you already know that. so you just have to deal with it.

  15. anonymous
    • one year ago
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    well thats what i don't know how to do

  16. freckles
    • one year ago
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    should be able to do integrate top-bottom

  17. freckles
    • one year ago
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    y=sin(x) is always above y=x^2-1 on the given interval

  18. anonymous
    • one year ago
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    wouldn't the integration include everything between the function and the red line too? |dw:1438891297985:dw|

  19. freckles
    • one year ago
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    so you want to include more area than shown?

  20. freckles
    • one year ago
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    |dw:1438891485605:dw| you want this area

  21. freckles
    • one year ago
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    |dw:1438891497073:dw|not this

  22. anonymous
    • one year ago
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    no, im just saying that if you find the integral of the top area it will find this area |dw:1438891538121:dw| which includes some area that i don't want

  23. freckles
    • one year ago
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    so you aren't convinced the area is given by: the integration of the (top-bottom) on the given interval? I may be able to help you understand that if that is what you don't understand

  24. anonymous
    • one year ago
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    im just not clear on how you would find the area of the top precisely thats all

  25. freckles
    • one year ago
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    |dw:1438891880460:dw|

  26. freckles
    • one year ago
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    let me just start with this piece this is going to give you overlap

  27. freckles
    • one year ago
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    you want to subract out that one part

  28. IrishBoy123
    • one year ago
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    i am such a moron shift the whole thing up by 1 integrating between \(y = x^2 -1\) and \(y=sin x\), "magnitude-wise", is same as integrating between \(y = x^2\) and \(y = sin x + 1\) you will have same x intercepts too

  29. anonymous
    • one year ago
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    you can do that !!!!!

  30. IrishBoy123
    • one year ago
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    ask @freckles sorry @freckles i had one of those "duh" moments......

  31. freckles
    • one year ago
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    yeah and we should wind up with \[\int\limits_{-0.637}^{1.141 }(\text{ top } - \text{ bottom }) dx \\ \int\limits_{-0.637}^{1.141 }(\sin(x) - (x^2-1)) dx \\ \] but trying to think how to explain this in pieces... \[\int\limits_0^{1.41 } \sin(x)dx-\int\limits_c^{1.41} (x^2-1) dx-\int\limits_0^c (x^2-1)-\int\limits_{-.637}^0 (x^2-1)dx--\int\limits_{-.637}^{0} \sin(x) dx \\ \text{ as we see this gives us } \\ \int\limits_{-.637}^{1.41} \sin(x) dx-\int\limits_{-.637}^{1.141}(x^2-1) dx \\ \text{ which can be written the way \above } \\ \int\limits_{-.637}^{1.41} (\sin(x)-(x^2-1) )dx\] it is hard for me though to draw some of this overlapping stuff

  32. anonymous
    • one year ago
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    okay so then it would be \[\int\limits_{-.636}^{1.409}\sin(x) +1 - x^2\] right?

  33. freckles
    • one year ago
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    yes exactly you just do top - bottom

  34. anonymous
    • one year ago
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    im getting -.376

  35. freckles
    • one year ago
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    shouldn't be a negative number

  36. anonymous
    • one year ago
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    for got the +1 1.6702

  37. anonymous
    • one year ago
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    Thank you very much :D

  38. anonymous
    • one year ago
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    one sec

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