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virtus

  • 2 years ago

area between y=x^2 -2x -3 and y =3x-3

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  1. saifoo.khan
    • 2 years ago
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    Try graphing.

  2. virtus
    • 2 years ago
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    i did but i don't know how to calculate it

  3. Hermeezey
    • 2 years ago
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    Take the definite integral of one and subtract out the definite integral of the other.

  4. FoolForMath
    • 2 years ago
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    HINT: Area under curve, Integration.

  5. saifoo.khan
    • 2 years ago
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    We integrate using the limits.

  6. FoolForMath
    • 2 years ago
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    Nathan is your Mr Analysis, he might help you :D

  7. virtus
    • 2 years ago
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    LOL the i like the sarcasm fool for math

  8. FoolForMath
    • 2 years ago
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    Hey vitus, long time :D

  9. nbouscal
    • 2 years ago
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    Set the equations equal to learn where they intersect and which is above the other, then take the definite integral of the lower function subtracted from the higher function, between the limits that you found.

  10. virtus
    • 2 years ago
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    i mean do you have have to split it some how and then integrate

  11. nbouscal
    • 2 years ago
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    Graphing is a good quick way to learn the limits of integration and which equation is larger.

  12. Hermeezey
    • 2 years ago
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    fine the points of interception, those will be your bounds of integration.

  13. virtus
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=area+between+y%3Dx%5E2+-2x+-3++and+y+%3D3x-3

  14. virtus
    • 2 years ago
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    thats how the graph looks like, how do i account for the are under the graph

  15. saifoo.khan
    • 2 years ago
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    |dw:1338596987767:dw|

  16. nbouscal
    • 2 years ago
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    By looking at that graph you can see that the two functions intersect at x=0 and x=5, so those will be your limits of integration. They also show you that the linear function is above the quadratic function, so you subtract the quadratic function from the linear function, then integrate it from 0 to 5.

  17. virtus
    • 2 years ago
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    OHHHH is that it!!!! LOL SORRY, didn't know why i couldn't see that.

  18. virtus
    • 2 years ago
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    thanks =]

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