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anonymous

  • one year ago

Show by hand how to find an anti-derivative to -t E^t^2

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  1. SolomonZelman
    • one year ago
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    u sub

  2. SolomonZelman
    • one year ago
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    u=t²

  3. jim_thompson5910
    • one year ago
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    hint: u-subsitution u = t^2, so du = 2t*dt

  4. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle\int\limits_{~}^{~}f'(x)\times e^{f(x)}~dx}\) SUBSTITUTION: `u=f(x) du=f'(x) dx` \(\large\color{black}{\displaystyle\int\limits_{~}^{~}e^{u}~du=e^u+C=e^{f(x)}+C}\) (this is an abstract case for some differentiable f(x) )

  5. anonymous
    • one year ago
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    u =t^2 du = 2t dt -E^u du = ...

  6. SolomonZelman
    • one year ago
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    well, not exactly

  7. anonymous
    • one year ago
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    I suppose (-1) can be pulled out

  8. nincompoop
    • one year ago
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    show by hand...

  9. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle\int\limits_{~}^{~}-t\times e^{t^2}~dt}\) \(\large\color{black}{\displaystyle-\int\limits_{~}^{~}t\times e^{t^2}~dt}\) \(\large\color{black}{\displaystyle u=t^2}\) \(\large\color{black}{\displaystyle du=2t~du~~~~\rightarrow~~~~\frac{1}{2}du=t~dt}\) \(\large\color{black}{\displaystyle-\frac{1}{2}\int\limits_{~}^{~} e^{u}~du}\)

  10. SolomonZelman
    • one year ago
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    like that - same as you said, BUT you got the 1/2 there (why? I have wrote why in my post)

  11. anonymous
    • one year ago
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    and that 2 has to be removed I think.. du = 2t dt du/2 = 2t dt/2 du/2 = t dt ???

  12. anonymous
    • one year ago
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    thanks solomon...

  13. SolomonZelman
    • one year ago
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    okay, can you now tell me what your antiderivative would be?

  14. SolomonZelman
    • one year ago
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    (antiderivative of e^x is just e^x ... (well, +C) )

  15. anonymous
    • one year ago
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    -1/2 E^t^2

  16. SolomonZelman
    • one year ago
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    yes, with +C

  17. anonymous
    • one year ago
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    oh yes.. +C

  18. SolomonZelman
    • one year ago
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    One of my past professors took off most of the partial credit for +C. He explained that by saying that you gave only 1/∞ of all possible answers.

  19. anonymous
    • one year ago
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    thanks, nicely explained.

  20. anonymous
    • one year ago
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    my professors have been really good with that.

  21. SolomonZelman
    • one year ago
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    Enjoy it any time I'm on. Although sometimes I will be gone. I wish good luck to you in all, Yes, that's the way! lets make it roll 0~0

  22. anonymous
    • one year ago
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    they had a grumble about how that happened to them.

  23. SolomonZelman
    • one year ago
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    If you have further questions, then u know - whenever I'm.... (or, there are many other math peeps here) Just always watch out how you solve for du in your substitution ....

  24. anonymous
    • one year ago
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    actually if I had to use FTC with something like this equation.. for limits of [a,b] would they expect me to put +C on that too? say... -1/2 E^b^2 - (-1/2 E^a^2)

  25. SolomonZelman
    • one year ago
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    wait, what does FTC stand for?

  26. anonymous
    • one year ago
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    fundamental theorem calculus

  27. SolomonZelman
    • one year ago
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    oh.

  28. SolomonZelman
    • one year ago
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    With limits of integration, +C ?

  29. anonymous
    • one year ago
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    I suppose it cancels out ...

  30. anonymous
    • one year ago
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    unless there's some kind of initial condition

  31. SolomonZelman
    • one year ago
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    If you are integrating a definite integral, you don't have the +C, because when you are integrating a definite integral you are finding the (numerical) area under the curve over a specific interval. And you are not finding the family of functions. So they say I guess: \(\large\color{black}{\displaystyle\int\limits_{a}^{b}f(x)~dx=\left( f(x)+C\right){\LARGE |}^{x=b}_{x=a}}\) (where F'=f)

  32. SolomonZelman
    • one year ago
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    and yes C cancel's out, but it shouldn't be there in a first place.

  33. SolomonZelman
    • one year ago
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    +C is a family of antiderivative functions F(x) (and this family of F(x)+C is a set of functions that has a derivative of f) when finding area under the corve over some interval I , that is not a family of functions you are finding - that is an area calculation.

  34. SolomonZelman
    • one year ago
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    Not that the result would differ if you have +C there, since it will cancel, but as far as the concept goes +C, again, should not be there)

  35. anonymous
    • one year ago
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    gotcha.. makes sense

  36. SolomonZelman
    • one year ago
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    Alrighty:)

  37. anonymous
    • one year ago
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    laterzs

  38. anonymous
    • one year ago
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    thanks again

  39. SolomonZelman
    • one year ago
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    Anytime

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