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anonymous

  • 5 years ago

whats the anti derivative of 1/(x^3)?? i have an answer but shows its wrong//// help?

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  1. anonymous
    • 5 years ago
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    my answer is x^-2/2

  2. anonymous
    • 5 years ago
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    When doing integrals with a power on the bottom, its often helpful to rewrite it like so anti derivative of x^-3 So lets use the general rule for integration, -3 + 1 = -2 So the anti derivative would be x^(-2) / -2 or in proper form 1 / 2*x^2

  3. anonymous
    • 5 years ago
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    I'm sorry, I mistyped that. The answer is -2 / x^2

  4. amistre64
    • 5 years ago
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    x^-3 -> x^-4/4: 1/(4x^4)

  5. anonymous
    • 5 years ago
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    Wait I keep messing that up.... 1 / -2*x^2

  6. amistre64
    • 5 years ago
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    hah!!... i missed it too lol

  7. amistre64
    • 5 years ago
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    -1/2x^2 is right ;)

  8. anonymous
    • 5 years ago
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    thanks you haha ... so how did you get that>?

  9. amistre64
    • 5 years ago
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    its the reverse of the power rule of a derivative.

  10. anonymous
    • 5 years ago
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    got it .. thanks

  11. amistre64
    • 5 years ago
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    \[D(X^n) = n * X^{n-1} \rightarrow D(x^5) = 5x^4\]

  12. dumbcow
    • 5 years ago
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    you do it correctly you just forgot to include the negative when you divided by the exponent

  13. amistre64
    • 5 years ago
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    \[\int\limits_{} X^n dx \rightarrow \frac{X^{n+1}}{n+1}\]

  14. anonymous
    • 5 years ago
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    ahh i got it ,.,.,.

  15. anonymous
    • 5 years ago
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    thanks

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