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anonymous

  • 5 years ago

how would you evaluate the definite integral of ((1+3x)/x^2)dx between 3 and 1?

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  1. amistre64
    • 5 years ago
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    split the fraction into two parts with common denominators

  2. amistre64
    • 5 years ago
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    1/x^2 + 3x/x^2 (S) x^-2 + 3x^-1 dx | [1,3]

  3. amistre64
    • 5 years ago
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    -1/x + 3 ln(x)

  4. amistre64
    • 5 years ago
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    -1+0 = -1 (-1/3 + ln(27)) - (-1)

  5. amistre64
    • 5 years ago
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    3.9625 there abouts if I did it right :)

  6. anonymous
    • 5 years ago
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    that is! thanks :D can you do this question? how do u find the antiderivative of sqrt (sinxcosx)? i need to find the volume of the soilid when the given graph is roated around the x-axis.

  7. amistre64
    • 5 years ago
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    hmmmm......maybe :)

  8. anonymous
    • 5 years ago
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    oh and you need to find the volume between 0 and pie/2

  9. amistre64
    • 5 years ago
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    change it to exponent stuff... (S) sin^(1/2) cos(1/2) dx

  10. amistre64
    • 5 years ago
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    let u = sin, du = cos dx might be useful

  11. anonymous
    • 5 years ago
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    oh ok. so intergration by parts.

  12. amistre64
    • 5 years ago
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    dx = du/cos yeah, im still new at the integrating, but I got some skill :)

  13. amistre64
    • 5 years ago
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    u^(1/2) cos^(1/2) -------- gives us what? cos

  14. amistre64
    • 5 years ago
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    im trying change of variable first :)

  15. anonymous
    • 5 years ago
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    actually you know what because i'm finiding the volume, i actually only need to find the antiderivative of sinxcosx

  16. amistre64
    • 5 years ago
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    really?...... but I was doing so good :)

  17. amistre64
    • 5 years ago
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    u= sin du = cosx dx (S) u du -> (sin^2)/2

  18. amistre64
    • 5 years ago
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    there should be punctuation in there somehere to clear things :)

  19. amistre64
    • 5 years ago
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    and if its bounded you dont need the +C

  20. anonymous
    • 5 years ago
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    that still doesn't make sense. if you make u=sinx, then du=cos x dx and dv=cos x and then v=sinx. we're back to where we started...?

  21. amistre64
    • 5 years ago
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    if its just the integral of sincos right?

  22. anonymous
    • 5 years ago
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    yea. and the equation is uv - integral of v du, when u integrate by parts.

  23. amistre64
    • 5 years ago
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    step back to a time when you were first learning this..... they probably went thru the easier stuff first like u-substitution and change of variables

  24. amistre64
    • 5 years ago
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    if we can get it into the form: (S) u du then thats all we need to go, no "by parts" or nothing. (S) sin cos dx ^ ^^^ (S) u du

  25. amistre64
    • 5 years ago
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    (S) u du -> (u^2)/2 (sin^2(x))/2 derive this and we get back to sin cos

  26. anonymous
    • 5 years ago
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    oh. lol. right. yea i get it...oh man. aha.

  27. amistre64
    • 5 years ago
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    you sure you dont need sqrt(sin cos) ??

  28. anonymous
    • 5 years ago
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    k, well the question is, find the volume of the solid obtained when the given region y=sqrt (sinxcosx)dx is roatated around the x-axis.

  29. anonymous
    • 5 years ago
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    and since the formula for volume is v= pie (s) f(x)^2 dx i can cancel the sqrt right?

  30. amistre64
    • 5 years ago
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    which is found by pi (S) [f(x)]^2 dx right?

  31. amistre64
    • 5 years ago
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    yes.... sounds good :)

  32. anonymous
    • 5 years ago
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    what answer did you get? if you have to find it between pi/2 and 0?

  33. amistre64
    • 5 years ago
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    lets see: sin(x)^2 ------ sin(0) = 0 so thats pointless 2 sin (pi/2) = 1 so I get 1/2 as the volume between 0 and 1

  34. anonymous
    • 5 years ago
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    i'm lame. my calculator was in degree mode...but i should have been able to figure it out anyways on paper.

  35. amistre64
    • 5 years ago
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    lol..... yeah :) these are basic angles and stuff from trig class :)

  36. anonymous
    • 5 years ago
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    i hate trig, i just type everything into my calc. ahha.

  37. amistre64
    • 5 years ago
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    i made it thru 3 semesters of math without no calculator :)

  38. amistre64
    • 5 years ago
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    last test they had odd angles that aint inthe "standard" tables.... broke my streak

  39. anonymous
    • 5 years ago
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    ahah. well. thanks a bunch.

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