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monroe17 Group Title

someone help? (:

  • one year ago
  • one year ago

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  1. monroe17 Group Title
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    \[\int\limits_{-2}^{0} (\frac{ 1 }{ 2 }t^4+\frac{ 1 }{ 4 }t^3-t)dt\]

    • one year ago
  2. nubeer Group Title
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    Hint. \[\int\limits_{?}^{?} t ^{4} =\frac{ t ^{4+1} }{ 4+1}\]

    • one year ago
  3. monroe17 Group Title
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    I honestly don't even know how to solve or begin to solve for it.. can you help me step by step? I have 7 more problems like this and I'd like to do them all myself.. But I need help solving and the steps to do so.

    • one year ago
  4. Shadowys Group Title
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    are you familiar with the general formula of integration?

    • one year ago
  5. monroe17 Group Title
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    I'm familiar with... \[\int\limits_{a}^{b} f(x)dx=F(x)|_{a}^{b} = F(b)-F(a)\]

    • one year ago
  6. nubeer Group Title
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    ya this is done when we apply the limits but do you know how we do integration?

    • one year ago
  7. monroe17 Group Title
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    I don't.

    • one year ago
  8. Shadowys Group Title
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    the general formula for integration of \(\int x^n dx = \frac{x^n+1}{n+1} +C\)

    • one year ago
  9. nubeer Group Title
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    |dw:1355140748339:dw|

    • one year ago
  10. winterfez Group Title
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    |dw:1355140890023:dw|

    • one year ago
  11. Shadowys Group Title
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    @winterfez i believe that's \(\left[ \frac{1}{2} \frac{t^5}{5}+\frac{1}{4} \frac{t^4}{4}- \frac{t^2}{2} \right]^0_{-2}\)

    • one year ago
  12. winterfez Group Title
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    \[(\frac{ 1 }{ 2 }\frac{ 0 }{ 5 }+\frac{ 1 }{ 4 }\frac{ 0^{4} }{ 4 }-\frac{ 0 }{ 2 })-(\frac{ 1 }{ 2 }\frac{ (-2)^5 }{ 5 }+\frac{ 1 }{ 4 }\frac{ (-2)^4 }{ 4 }-\frac{ (-2)^{2} }{ 2 })\]

    • one year ago
  13. winterfez Group Title
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    now use your calculator to punch in those suckers

    • one year ago
  14. monroe17 Group Title
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    can you explain the [...]_{-2}^{0}

    • one year ago
  15. Shadowys Group Title
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    ∫abf(x)dx=F(x)|ba=F(b)−F(a) <--this is why.

    • one year ago
  16. Shadowys Group Title
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    from your original post up there.

    • one year ago
  17. monroe17 Group Title
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    ohhh so i do apply that.. I just missed a step.

    • one year ago
  18. Shadowys Group Title
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    yup :) i think you've got it.

    • one year ago
  19. monroe17 Group Title
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    thank you guys! (: you're amazing!

    • one year ago
  20. Shadowys Group Title
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    you're welcome :)

    • one year ago
  21. monroe17 Group Title
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    I got 21/5.. does that seem right @Shadowys

    • one year ago
  22. Shadowys Group Title
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    -21/5?

    • one year ago
  23. monroe17 Group Title
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    err... let me recheck my work.

    • one year ago
  24. Shadowys Group Title
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    sorry, my mistake. i typed wrong. lol no neg.

    • one year ago
  25. monroe17 Group Title
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    oh okay (: is that the answer though? because what happens with the .... +C?

    • one year ago
  26. Shadowys Group Title
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    well. that's for indefinite integrals because you don't know where is the limit. the C is gone when it is definite.

    • one year ago
  27. monroe17 Group Title
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    oh okay (: so 21/5 it is?

    • one year ago
  28. Shadowys Group Title
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    yup! :)

    • one year ago
  29. monroe17 Group Title
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    thanks again!

    • one year ago
  30. Shadowys Group Title
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    you're welcome :)

    • one year ago
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