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Loser66

  • one year ago

\[\int_{-3}^{3} |x+1|dx\] Please, help

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  1. jtvatsim
    • one year ago
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    Integrate separately on parts of the domain that are convenient is the quick tip. :)

  2. Loser66
    • one year ago
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    Show me, please.

  3. jtvatsim
    • one year ago
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    Or, just draw it and find the area. :)

  4. jtvatsim
    • one year ago
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    Will do, I'll do the drawing method since it's more clever.

  5. Loser66
    • one year ago
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    I show you my confusion.

  6. jtvatsim
    • one year ago
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    |dw:1433638297626:dw|

  7. Loser66
    • one year ago
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    no confuse, hehehe.. I am sorry. I posted the wrong question. But I would like to know how to solve it traditionally

  8. jtvatsim
    • one year ago
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    No worries. :) You should be able to see from the graph that the answer is the area of two triangles. But I will do the traditional (algebraic nastiness) approach as well.

  9. Loser66
    • one year ago
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    I meant integrate separately..........

  10. jtvatsim
    • one year ago
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    First, we need the definition of what |x + 1| means: |x + 1| = { x + 1, for x >= -1; -(x + 1), for x < -1. This is the usual, "make the number positive" rule we remember in our heads expressed algebraically.

  11. jtvatsim
    • one year ago
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    It may take a bit of pondering to fully see that. :)

  12. jtvatsim
    • one year ago
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    Then, this naturally gives us two convenient domains. (-infinity, -1) and (-1, +infinity). The split point is at x = -1.

  13. jtvatsim
    • one year ago
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    We want (-3, 3) so we break this into the domains (-3, -1), and (-1, 3).

  14. jtvatsim
    • one year ago
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    Then we integrate using the definition of absolute value.

  15. jtvatsim
    • one year ago
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    \[\int\limits_{-3}^3 |x + 1| \ dx = \int\limits_{-3}^{-1} -(x+1) \ dx + \int\limits_{-1}^3 x+1 \ dx\]

  16. jtvatsim
    • one year ago
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    The integration can then be carried out as usual.

  17. Loser66
    • one year ago
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    Got you. Thanks a lot

  18. jtvatsim
    • one year ago
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    No problems! :)

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