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anonymous

  • one year ago

how do you integrate sqrt(8x^2)

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  1. anonymous
    • one year ago
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    |dw:1442604743169:dw|

  2. IrishBoy123
    • one year ago
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    that's\[\int \sqrt{8} \, \sqrt{x^2} \, dx\]

  3. anonymous
    • one year ago
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    how would do this using u subs

  4. IrishBoy123
    • one year ago
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    \[\sqrt{8} = ??\] \[\sqrt{x^2} = ??\]

  5. anonymous
    • one year ago
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    |dw:1442605850479:dw|

  6. anonymous
    • one year ago
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    |dw:1442605975556:dw|

  7. IrishBoy123
    • one year ago
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    yes, i would see it that way initially but the problem is i have a feeling that \(\sqrt{x^2}\) is a trap let me ask some others. sorry about this, but i think it might be worthwhile @ganeshie88

  8. IrishBoy123
    • one year ago
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    @ganeshie8

  9. anonymous
    • one year ago
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    ok

  10. IrishBoy123
    • one year ago
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    @Empty

  11. Empty
    • one year ago
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    Yeah I definitely think it's a trap :P

  12. anonymous
    • one year ago
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    i try solving this using the u substitution where u=8x^2, du=16x dx and now the problem is i have an x hanging with 16

  13. Empty
    • one year ago
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    I don't think a substitution is the right choice here. I think @IrishBoy123 has the right strategy here, separate the constant out front like this: \[\sqrt{8} \int \sqrt{x^2}dx\] The problem is specifically this subtle fact: \[\sqrt{x^2} \ne x\] Here's the proof, let x be some negative number such as -2 why not that's a possible value x can take: \[\sqrt{(-2)^2} = \sqrt{2^2} \ne -2\] So really what we have is the absolute value function! This is kind of tricky to deal with, but easier if you have bounds on your integral: \[\int \sqrt{x^2} dx = \int |x| dx = \int_{-a}^0 -x dx + \int_0^b x dx \] A picture will hopefully clarify this a bit! |dw:1442607260478:dw|

  14. anonymous
    • one year ago
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    Thank you very much

  15. Empty
    • one year ago
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    Cool so was that all no questions?! :O

  16. anonymous
    • one year ago
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    there are lol ill try to do them myself and if i cant ill come back for help

  17. IrishBoy123
    • one year ago
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    ah! thanks a million, @Empty

  18. anonymous
    • one year ago
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    how do i integrate this now |dw:1442608670564:dw|

  19. anonymous
    • one year ago
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    @ganeshie8

  20. anonymous
    • one year ago
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    @Empty

  21. IrishBoy123
    • one year ago
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    @magepker728 make a new thread for this new question this one is a tan sub [or maybe, just for fun, a hyperbolic: \(cosh^2(x) − sinh^2(x) = 1 \)].

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