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anonymous

  • one year ago

Simple calculus question

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  1. nincompoop
    • one year ago
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    So simple it has no answer

  2. anonymous
    • one year ago
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    Im stuck on a specific type of question, can someone help me?

  3. anonymous
    • one year ago
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  4. amistre64
    • one year ago
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    do you know the integration of sec(t) ?

  5. anonymous
    • one year ago
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    Why is that necessary?

  6. amistre64
    • one year ago
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    its not really, but its a good check ...

  7. anonymous
    • one year ago
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    Im not quite sure where sec come into this.

  8. amistre64
    • one year ago
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    sec and csc have similar derivatives i recall sec and then adjust for csc

  9. amistre64
    • one year ago
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    the integration of sec was well known prior to the founding of integration :)

  10. anonymous
    • one year ago
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    The derivative of csc i believe is -cscxcotx

  11. amistre64
    • one year ago
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    the derivative yes, but the integration? its a sneaky little trick

  12. amistre64
    • one year ago
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    \[\frac{csc}{1}*\frac{-csc-cot}{-csc-cot}\] \[\frac{-csc^2-csc~cot}{-csc-cot}\implies~ln(csc+cot)\]

  13. anonymous
    • one year ago
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    Im lost, simply how do you go from a derivative to an integral which is equal to the original function. Is it as simple as moving the derivative into the integral?

  14. anonymous
    • one year ago
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    It's the fundamental theorem of calculus \[\frac{ d }{ dx } \int\limits_{a}^{x} f(t) dt = f(x)\]

  15. amistre64
    • one year ago
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    we have 2 options, we can use the fundamental thrm: \[\int_{a}^{x}f'(t)dt=F(x)-F(a)\] and take the derivative \[\frac d{dx}\int_{a}^{x}f'(t)dt=f'(x)x'-f'(a)a'\]

  16. anonymous
    • one year ago
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    So the derivative of the integral of a function in that format is equal to that function.. so since i have the derivative do i find the anti-derivative and plug that into the integral?

  17. amistre64
    • one year ago
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    or we can work the integration, and then take the derivative, either way

  18. anonymous
    • one year ago
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    My course has recently worked with the fundamental therom so I assume thats the way they want me to solve it.

  19. amistre64
    • one year ago
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    you asked about which one it shouldbe, im just suggesting that in a pinch, you can work the long way if possible

  20. anonymous
    • one year ago
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    So... does that make the answer b or am i on the wrong route?

  21. anonymous
    • one year ago
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    It seems to me this problem is designed to be simple im just making it difficult.

  22. amistre64
    • one year ago
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    it is designed to make you apply the fundamental thrm yes

  23. anonymous
    • one year ago
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    Hey sorry, how did you get \[\frac{-\csc^2-\csc~\cot}{-\csc-\cot}\implies~\ln(\csc+\cot)\]

  24. anonymous
    • one year ago
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    Hmm?

  25. amistre64
    • one year ago
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    prolly an error in my head, thinkig to quick -ln(csc + cot) derives to csc like i said, im used to the sec form :)

  26. amistre64
    • one year ago
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    x should be the high range, so its either the first or last option to me

  27. anonymous
    • one year ago
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    Or b

  28. anonymous
    • one year ago
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    the negation flips the limits, correct?

  29. anonymous
    • one year ago
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    never mind

  30. anonymous
    • one year ago
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    I know its not c (duh). I dont believe it is a. However i dont know whether i put the antiderivtive of the derivitive into the intergral or the derivative into the integral

  31. anonymous
    • one year ago
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    IE, b or d

  32. amistre64
    • one year ago
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    -ln(csc(x)+cot(x)) + C derives to csc(x) since y = -ln(csc(x)+cot(x)) + C, i dont see why we would have a constant in the integration when dy/dx = csc(x)

  33. amistre64
    • one year ago
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    a or d is my thought

  34. anonymous
    • one year ago
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    the constant i believe has something to do with making the y value correct, they all have them so i think its correct.

  35. anonymous
    • one year ago
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    A or D was what i started stuck between XD

  36. anonymous
    • one year ago
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    im sorry b or d

  37. amistre64
    • one year ago
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    \[y=\int y'(t)~dt +C\] i see it now, these old eyes were placing it inside the dt

  38. anonymous
    • one year ago
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    XD thats a correct therom? so that would make it d?

  39. anonymous
    • one year ago
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    Essentially there youre using the intergral to find the antiderivative?

  40. amistre64
    • one year ago
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    lets do the long way and check it out \[y=\int_{a}^{x}csc(t)~dt+C=-ln(csc(x)+cot(x))+ln(csc(a)+cot(a))+C~\] when x=4, y=-9 \[-9+ln(csc(4)+cot(4))=ln(csc(a)+cot(a))+C\]

  41. amistre64
    • one year ago
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    im using the integral to determine that solution yes :)

  42. amistre64
    • one year ago
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    let a=4, and C=-9

  43. anonymous
    • one year ago
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    Thanks :) It really was simple XD

  44. anonymous
    • one year ago
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    I figured

  45. amistre64
    • one year ago
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    notice that by comparing like parts: ln(csc(4)+cot(4)) = ln(csc(a)+cot(a)) , let a=4 -9 = C ...

  46. amistre64
    • one year ago
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    another way to have viewed this, now that i have my bearings straight might be like this: 4 is in the domain element, it gets used as an input value ... it should be in the limit interval -9 is a range element, its not necessarily a part of the integrals domain limits

  47. amistre64
    • one year ago
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    but thats just a guess at what i see, i would not have determined that by just the FTC

  48. anonymous
    • one year ago
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    Thanks: i have actually solved this equation before (and whtat you are saying sounds farmilar) but i was dead tired and couldnnt remember how i did it.

  49. amistre64
    • one year ago
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    good luck :)

  50. anonymous
    • one year ago
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    not to butt in, but why isn't the answer obviously 4?

  51. anonymous
    • one year ago
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    Haha I was thinking the same thing, but anyway it was a nice question thanks @sccitestla and thanks @amistre64 :)

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