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anonymous

  • 5 years ago

Taking the derivative of an integral...

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  1. anonymous
    • 5 years ago
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    How do I take the derivative of the function \[g(x)=\int\limits_{x}^{x^2}(t^2-5)/(t^2+2)dt\]

  2. amistre64
    • 5 years ago
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    the derivative and an integral are inverse operations: thed cancel each other out and youd end up with: g'(x) = (t^2-5)/(t^2+2) ..... as far as I recollect

  3. anonymous
    • 5 years ago
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    How exactly does it cancel out?

  4. amistre64
    • 5 years ago
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    hmmm..... you might have a point, I failed to recognize the variables....

  5. anonymous
    • 5 years ago
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    So.. how would you do it?

  6. amistre64
    • 5 years ago
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    take the integral, and substitute for t with x...... then derive it back down..... but the actuall process eludes me :)

  7. amistre64
    • 5 years ago
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    can we get the integrand into a sum of terms?

  8. amistre64
    • 5 years ago
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    dunno if a trig substitution would be useful: t = sqrt(2) tan(a)

  9. anonymous
    • 5 years ago
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    Idk.. I'm just totally lost on this question. So I substitute the t's for x's?

  10. anonymous
    • 5 years ago
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    Or do I substitute the t^2 in?

  11. amistre64
    • 5 years ago
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    my limited knowledge on this would suggest that I find g(x) from the integrand first.... an exact HOW, is beyind me at the moment :)

  12. anonymous
    • 5 years ago
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    Lol ok. Thank you for your help though :)

  13. amistre64
    • 5 years ago
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    if I did it right in my thoguths here: the new integrand becomes: tan^2(a) -5/2 which looks easier to work with...

  14. anonymous
    • 5 years ago
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    nicole did you finish most of the packet? i didn't...i do have a much geater understanding about integrals though :)

  15. anonymous
    • 5 years ago
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    No.. I think I'm gonna stay after school.

  16. anonymous
    • 5 years ago
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    mother just told me she is picking me up. i have NO idea why. i want to stay too..

  17. amistre64
    • 5 years ago
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    i substituted: t = sqrt(2) tan(a) ; dt = sqrt(2) sec^2 2tan^2(a) -5 sqrt(2)sec^2(a) da ------------- ------------- 2(tan^2(a) + 1) 1 2tan^2(a) -5 sqrt(2)sec^2(a) da ------------- ------------- 2sec^2(a) 1 2tan^2(a) -5 sqrt(2) da ------------- ------- 2 1 2tan^2(a) -5 sqrt(2) da ---------- - --- ------- 2 2 1 sqrt(2) tan^2(a) - 5sqrt(2)/2 da is what I get actually, but it might be easier to integrate

  18. anonymous
    • 5 years ago
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    Just stay.. find a ride home.

  19. anonymous
    • 5 years ago
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    Wow.. That's a whole lot..

  20. amistre64
    • 5 years ago
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    or since the sqrt(2) is a constant, we can yank it out to the other side: sqrt(2) [S] tan^2(a) - 5/2 da

  21. amistre64
    • 5 years ago
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    myins here....now ill now what Im doing wrong ;)

  22. myininaya
    • 5 years ago
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    Let f(t)=(t^2-5)/(t^2+2) Then int(f(t), t=x..x^2)=F(x^2)-F(x) To take derivative of this. We have to use chain rule. Therefore g'(x)=2xf(x^2)-f(x)

  23. myininaya
    • 5 years ago
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    g'(x)=2x[(x^4-5)/(x^4+2)]-(x^2-5)/(x^2+2)

  24. myininaya
    • 5 years ago
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    any questions?

  25. myininaya
    • 5 years ago
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    we dont need to integrate to find the derivative of the integral

  26. amistre64
    • 5 years ago
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    ......thats what I thought :) but those x..x^2 things threw me for a loop

  27. myininaya
    • 5 years ago
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    lol where did you get all of those fans? I'm so jealous

  28. amistre64
    • 5 years ago
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    they were having a sale at pier1 ;)

  29. myininaya
    • 5 years ago
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    damn i missed it :(

  30. myininaya
    • 5 years ago
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    i think he lefted :(

  31. amistre64
    • 5 years ago
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    maybe, but the math remained lol

  32. myininaya
    • 5 years ago
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    true im going to go play call of duty now peace

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