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anonymous

  • one year ago

If f[x] = Integrate]t/0 4 Sin[x^3] what does it mean to 'give a clean formula' for f'[x] ? latex coming ...

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  1. anonymous
    • one year ago
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    \[f(x)= \int\limits_{0}^{t} 4 \sin (x^3) dx\]

  2. Astrophysics
    • one year ago
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    Looks like they want you to use fundamental theorem of calculus part 1

  3. anonymous
    • one year ago
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    would that not just be \[\int\limits_{0}^{t} f'(x) = f(t) - f(0)\] and there for f'(x) = 4 sin (x^3) Or is there more to it than that?

  4. Astrophysics
    • one year ago
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    Kind of all you have to do here is plug in t where x is

  5. anonymous
    • one year ago
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    ahhh thats what I did wrong

  6. anonymous
    • one year ago
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    thnx

  7. Astrophysics
    • one year ago
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    So we have \[\int\limits_{0}^{t} 4 \sin(x^3) dx = 4\sin(t^4)=f'(x)\]eas right

  8. Astrophysics
    • one year ago
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    easy*

  9. Astrophysics
    • one year ago
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    Np :-)

  10. anonymous
    • one year ago
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    had something like d/dx f[t] = f'[x]

  11. Astrophysics
    • one year ago
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    Sort of haha, basically taking the derivative of integral, which just cancels out the integral, if that makes sense

  12. Astrophysics
    • one year ago
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    It'll be rare for you to see fundamental theorem of calc part 1 after calc 1 xD

  13. Astrophysics
    • one year ago
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    But part 2 what you did first is everywhere

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