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anonymous

  • 5 years ago

cal 1 question.

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  1. anonymous
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    =cos(1+x+x^2)

  3. anonymous
    • 5 years ago
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    fundamental theorem of calc says: the derivative of the integral is the integrand. so in the simplest language the answer is the function being integrated, but be sure to replace t by x (since the integral is a function of x, not t) \[cos(1+t+t^2)\] easy right?

  4. anonymous
    • 5 years ago
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    oops i meant\[cos(1+x+x^2)\] sorry.

  5. anonymous
    • 5 years ago
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    but thats not the answer

  6. anonymous
    • 5 years ago
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    =/

  7. anonymous
    • 5 years ago
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    really?

  8. amistre64
    • 5 years ago
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    we need to find the integrand from -1 to x and then derive it back

  9. amistre64
    • 5 years ago
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    u = (1+x+x^2); du = 1+2x dx du/1+2x = dx maybe?

  10. anonymous
    • 5 years ago
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    i am sticking with my answer. unless they changed the fundamental theorem of calculus or unless i read the problem incorrectly.

  11. anonymous
    • 5 years ago
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    How come?! I am pretty sure it's the answer.

  12. amistre64
    • 5 years ago
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    the integrand changes in form and then you derive the change in form down to the answer, i think

  13. amistre64
    • 5 years ago
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    try it on a simpler problem; {S} 1+x^2 dx ; [-1,x] x + x^3/3 gets what?

  14. amistre64
    • 5 years ago
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    x^2 +x^4/3 -(-1 + -1/3)

  15. anonymous
    • 5 years ago
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    i am afraid i disagree. i read the problem again and it clearly asks for the derivative of the integral. if \[F(x)=\int_a^x f(t)dt\] then \[F'(x)=f(x)\]

  16. amistre64
    • 5 years ago
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    Dx( (1/3)x^4 +x^2 + 4/3) (4/3) x^3 +2x right?

  17. anonymous
    • 5 years ago
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    i am curious as to what the answer is supposed to be. an integral of the form asked about is not an anti-derivative, it is a reimann sum, aka the limit of a sum. the derivative of such an animal is in fact the integrand. not too hard to prove this. that is the reason why you get to use anti-derivatives to compute integrals.

  18. anonymous
    • 5 years ago
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    i know i got the same answer BUT thats not the answer. the software only shows when an aswer is correct so i don not know the correct answer

  19. anonymous
    • 5 years ago
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    blame the software. what exactly did you type in?

  20. anonymous
    • 5 years ago
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    and one more thing: why aren't we differentiating 1+ t+ t^2

  21. anonymous
    • 5 years ago
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    use the newton-leibniz theorem

  22. anonymous
    • 5 years ago
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    you are not asked for the derivative of the INTEGRAND. you are asked for the derivative of the INTEGRAL.

  23. anonymous
    • 5 years ago
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    I checked with a software, I got the same answer. Trust me the answer that satellite73 said earlier is right. This is the fundamental theorem of calculus; the first theorem you study in integration.

  24. anonymous
    • 5 years ago
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    \[F(x)=\int_a^x f(t)dt\] is a function of x, not of t.

  25. anonymous
    • 5 years ago
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    derivative of f(t)dt integral from f1(x) to f2(x) is f(f1)f1' - f(f2)f2'

  26. anonymous
    • 5 years ago
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    okay probably the problem with teh software.

  27. anonymous
    • 5 years ago
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    and (at the risk of repeating myself) the derivative is \[f(x)\]

  28. amistre64
    • 5 years ago
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    so was i right in my thought? :) integrate than derive the result

  29. anonymous
    • 5 years ago
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    hey btw why arent we differentiating 1+t+t^.5

  30. anonymous
    • 5 years ago
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    i told u guys

  31. amistre64
    • 5 years ago
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    {S} cos(1+t+t^2) dt ; [-1,x] = what?

  32. anonymous
    • 5 years ago
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    the answer is cos(1 +x + x^2)

  33. anonymous
    • 5 years ago
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    amistre, i am afraid that is not correct. the definition of the integral \[\int_a^x f(t)dt\] is not an "anti-derivative" and in fact \[\int_{-1}^x x + x^2 dx\] doesn't make any sense. you are using x both as the upper limit of integration, and as the 'dummy variable' in the integrand.

  34. amistre64
    • 5 years ago
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    oh....

  35. amistre64
    • 5 years ago
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    so i shoulda used another variable to test with then i spose :) thnx

  36. anonymous
    • 5 years ago
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    Here, this is the final answer after evaluating the integral and then taking its derivative: http://bit.ly/lIvQ5I

  37. anonymous
    • 5 years ago
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    \[\int\limits_{g(x)}^{h(x)}f'(t)dt = f[h(x)] - f[g(x)\] now if u differentiate it \[f'[h(x)]h'(x) - f'[g(x)]g'(x)\]

  38. anonymous
    • 5 years ago
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    you seem to be asking for a closed form of the integral, which then you wish to differentiate. i have no idea what the closed form of the integral is, nor do i care. it probably doesn't even have one. the probability that you can find an 'anti derivative' for a randomly chosen function is 0. we only work with special cases where we show off by saying "wow look, i found a function whose derivative is the one given!" but in general this is impossible to do. most integration is done using numerical methods

  39. anonymous
    • 5 years ago
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    this is called newton-leibniz rule...for derivatives of integral forms

  40. anonymous
    • 5 years ago
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    GUYS IM EXPLAINING THE ANSWER 2 U..IS ANYONE LISTENING

  41. anonymous
    • 5 years ago
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    rsaad did u get the answer????

  42. anonymous
    • 5 years ago
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    yea i got it =) thnx

  43. anonymous
    • 5 years ago
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    did u understand the working i explained?

  44. anonymous
    • 5 years ago
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    yes i got that BUT as far as i can remember, in ftoc 1, u also differentiate wats inside the finction.....cuz of chain rule

  45. anonymous
    • 5 years ago
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    bt weve done that havent we...ive explained above..see..ive proved the rule for any general function

  46. anonymous
    • 5 years ago
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    him1618 you are certainly correct IF f is an anti derivative of f' etc. the point is that the derivative of the integrand is the integral irrespective of whether or not you can find an anti-derivative. you are stating a (correct) theorem which assumes you have an anti-derivative of f. in this case we do not, nor does it matter. if you look at the attachment from Anwar A you will see that mathematica didn't even find an anti-derivative. that weird looking thing after the "dx" on the first line includes C and S which are in fact themselves integrals.

  47. anonymous
    • 5 years ago
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    oops typo. i meant to say that the derivative of the integral is the integrand, not the other way around.

  48. anonymous
    • 5 years ago
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    yes of course having an antiderative for a function is prerequisite for doing this..bt i guess the question setter didnt care to go deep enough to check whether this function ws even integrable or not...so i guess they just want rsaad2 to solve the question...u have a point satellite..but i think the setter hasnt gone that deep with his question...so he just needs answers..;)

  49. anonymous
    • 5 years ago
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    people just explain as to why the chain rule does not apply to whats in th funcyion that is whats in cos (.............)?

  50. anonymous
    • 5 years ago
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    were using the chain rule in the second step here saad

  51. anonymous
    • 5 years ago
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    yes. so what abt wats in the cos function

  52. anonymous
    • 5 years ago
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    first we are integrating a function...with limits which are variable wrt another variable..and then we differentiate using chain rule

  53. anonymous
    • 5 years ago
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    ok.

  54. anonymous
    • 5 years ago
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    oh the chain rule does apply but just remember that \[\int_a^x f(t)dt\] is a function of x, not a function of t. so a composite function would look something like \[\int_a^{h(x)} f(t)dt\]

  55. anonymous
    • 5 years ago
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    se wht i wrote above

  56. anonymous
    • 5 years ago
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    exactly!

  57. anonymous
    • 5 years ago
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    um =S

  58. anonymous
    • 5 years ago
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    plz explain it a bit more

  59. anonymous
    • 5 years ago
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    so the derivative of \[\int_{-1}^{x^3} cos(1+t+t^2)dt\] would be \[cos(1+x+x^2)\times 3x^2\]

  60. anonymous
    • 5 years ago
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  61. anonymous
    • 5 years ago
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    see this file..ull understand

  62. anonymous
    • 5 years ago
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    why not cos(1+x+x^2) x (2x+1) ?

  63. anonymous
    • 5 years ago
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    throughout this operation we never differentiate the cos function..do u get it?

  64. anonymous
    • 5 years ago
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    yea i get the cos part but why not wats inside it

  65. anonymous
    • 5 years ago
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    damn i made a mistake. sorry. i will correct it.

  66. anonymous
    • 5 years ago
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    we integrate the cos fn with two separate functions of x as limits first, and then we differentiate the integral of the cos fn, which gives us the function back, note that we are differentiating the inside part, but after replacing t by f(x) and g(x) in ur case x and -1

  67. anonymous
    • 5 years ago
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    the chain rule says that the derivative of \[F(G(x))\] is \[F'(G(x))\times G'(x)\]

  68. anonymous
    • 5 years ago
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    if i put \[F(x)=\int_{-1}^x cos(1+t+t^2)dt\] then its derivative is \[cos(1+x+x^2)\]

  69. anonymous
    • 5 years ago
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    ok

  70. anonymous
    • 5 years ago
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    thanks everyone!

  71. anonymous
    • 5 years ago
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    no prob mate

  72. anonymous
    • 5 years ago
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    but if i write \[\int_{-1}^{x^3} cos(1+t+t^2)dt\] then its derivative is \[cos(1+x^3+(x^2)^3)\times 3x^2\]

  73. anonymous
    • 5 years ago
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    right

  74. anonymous
    • 5 years ago
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    the derivative of the outside function evaluated at the inside function times the derivative of the inside function. i made a mistake earlier.

  75. anonymous
    • 5 years ago
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    yeah..now righter lol;)

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