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Abhisar

  • one year ago

Integration Help! Please guide through the steps \(\huge \sf \int\limits_{0}^{4} (6t-8)(3-8t+3t^2).dt\)

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  1. imqwerty
    • one year ago
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    ok 1st we need to open the brackets :)

  2. imqwerty
    • one year ago
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    wait....the method with which i did it was really long but i got the answer 176 wait lemme think of a shorter one :)

  3. arindameducationusc
    • one year ago
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    hmmm. I can do this... but as @imqwerty is doing. I don't need to interfere. If he has trouble then I will jump in @Abhisar

  4. SolomonZelman
    • one year ago
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    u substitution

  5. Abhisar
    • one year ago
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    You mean like this \(\sf ∫u v dx = u∫v dx −∫u' (∫v dx) dx\)

  6. SolomonZelman
    • one year ago
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    Well, we know that if you have something like an abstract case below: \(\large\color{black}{ \displaystyle \int_{a}^{b} \color{red}{f'(x)}~{\rm G}\left(\color{red}{f(x)}\right)dx }\) f'(x) is just some other function that is derivative in relation to the f(x) part, then you set \(u=f(x)\) \(du=f'(x)~du\) \(x=a~~~\rightarrow~~u=f(a)\) \(x=ba~~~\rightarrow~~u=f(b)\) and you would then have: \(\large\color{black}{ \displaystyle \int_{f(a)}^{f(b)} {\rm G}\left(\color{red}{u}\right)du }\)

  7. SolomonZelman
    • one year ago
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    I meant \(du=f'(x)~dx\)

  8. SolomonZelman
    • one year ago
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    should i give ordinary, reg. example?

  9. Abhisar
    • one year ago
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    Yes, please c:

  10. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \int_{2}^{4} 2x\cdot (x^2-3)^{100}dx }\) you don't want to epand, do you? O~O you set \(u=x^2-3\) and then \(du=2x~dx\) (see how nice it is that you have that "2x" part in the integral) and when x=2, u=\((\color{red}{2}^2-3)=1\) (because u in relation to x is x²-3) and when x=4, u=\((\color{red}{4}^2-3)=13\) So you get: \(\large\color{black}{ \displaystyle \int_{1}^{13} (u)^{100}du }\)

  11. SolomonZelman
    • one year ago
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    and then the rest in that example is obvious (note, that I change the limits of integration equivalently, so you don't need to sub back the x)

  12. arindameducationusc
    • one year ago
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    I did it in a different way..... (I dont understand why it should be wrong) |dw:1440428321372:dw| I multiplied the brackets

  13. Abhisar
    • one year ago
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    Please, wait....Let me digest it c:

  14. SolomonZelman
    • one year ago
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    |dw:1440428430105:dw|

  15. SolomonZelman
    • one year ago
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    take your time

  16. arindameducationusc
    • one year ago
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    |dw:1440428440544:dw|

  17. ganeshie8
    • one year ago
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    why is \(u\) substitution dropped ?

  18. SolomonZelman
    • one year ago
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    what do you mean ganeshie?

  19. SolomonZelman
    • one year ago
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    dropped, as "dropped into the thread" - saying, why is it proposed? or, "dropper" as left out?

  20. SolomonZelman
    • one year ago
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    dropped*

  21. ganeshie8
    • one year ago
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    im asking why are you expanding that out instead of simply using \(u\) substitution..

  22. SolomonZelman
    • one year ago
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    Well, arinda.. propsed that approach as it is also a posibility (in this case) for those that aren't familir with u sub. I just say that derivative of 3t^2-8t+3 by so I said u sub right off

  23. SolomonZelman
    • one year ago
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    just *see* (typos, oh my)

  24. arindameducationusc
    • one year ago
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    One question,

  25. SolomonZelman
    • one year ago
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    Abhisar, if you want another example let me know ...

  26. arindameducationusc
    • one year ago
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    Without u substitution... it can be solved this way right?

  27. ganeshie8
    • one year ago
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    Ahh okay nice :) its just that i felt this problem is begging for \(u\) substitution... expanding it out is somewhat pain...

  28. SolomonZelman
    • one year ago
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    Yes, arinda it is possible, certainly:)

  29. arindameducationusc
    • one year ago
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    okay.... :D

  30. SolomonZelman
    • one year ago
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    not that i would abide by that approach in such case.

  31. Abhisar
    • one year ago
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    One min, let me try this method on the original problem.

  32. SolomonZelman
    • one year ago
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    Sure, go ahead:) \(!^\infty\)

  33. arindameducationusc
    • one year ago
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    I don't know about u substitution, could you guyz teach me?

  34. SolomonZelman
    • one year ago
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    this is not your question, apologize to remind you-;(

  35. SolomonZelman
    • one year ago
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    You can make your own tho' there are plenty people (just on this post, certainly on the site) that can help you much better than little me

  36. arindameducationusc
    • one year ago
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    ya @SolomonZelman but thanks to @Abhisar to raise this question, I got to know about u substitution!

  37. Abhisar
    • one year ago
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    Ok, so applying u substitution method on the original problem, I can choose any one as u. Correct?

  38. SolomonZelman
    • one year ago
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    Well, you can choose anything technically, but I really don't promise it will be easier if you choose not the ideal one

  39. SolomonZelman
    • one year ago
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    You know that when you substitute u=f(x) in any case, then whatever the derivative of this f(x) is (we will reasonably call it f'(x)) then the new du replaces "f'(x) and dx"

  40. Abhisar
    • one year ago
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    ok, let's start with \(\sf 3-8t+3t^2\) as u, then \(du = -8+6t\).

  41. SolomonZelman
    • one year ago
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    Tell me what is the derivative (with respect to t) of \(3t^2-8t+3\) ?

  42. SolomonZelman
    • one year ago
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    yes, correct LOL

  43. Abhisar
    • one year ago
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    c:

  44. Abhisar
    • one year ago
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    Now, what?

  45. SolomonZelman
    • one year ago
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    and knowing that u=3t²-8t+3 you can find the corresponding limits of integration when t=0, u=what? when t=4, u=what?

  46. arindameducationusc
    • one year ago
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    I am watching :D Nice.....

  47. Abhisar
    • one year ago
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    and knowing that u=3t²-8t+3 you can find the corresponding limits of integration when t=0, u=3 when t=4, u=19

  48. SolomonZelman
    • one year ago
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    yes, nice

  49. Abhisar
    • one year ago
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    ok, now? c:

  50. SolomonZelman
    • one year ago
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    So, what you do then is: \(\large\color{black}{\displaystyle\int\limits_{0}^{4}\color{red}{(6t-8)}(3-8t+3t^2)\color{red}{dt}}\) the red part is replaced by du and the `3-8t+3t²` piece is your you u, so you put that in too. Also, put in your new limits of integration that correspond to your "u". \(\large\color{black}{\displaystyle\int\limits_{3}^{19}(u)\color{red}{du}}\)

  51. SolomonZelman
    • one year ago
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    the `3-8t+3t²` is your \(\cancel{\rm you}\) u

  52. SolomonZelman
    • one year ago
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    did i confuse you by typing too much? (sorry for this habit then)

  53. Abhisar
    • one year ago
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    No it's actually helping me. Wait a sec.

  54. SolomonZelman
    • one year ago
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    sure

  55. SolomonZelman
    • one year ago
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    if you don't understand something, then don't hesitate to ask...

  56. Abhisar
    • one year ago
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    So I write, \(\large\color{black}{\displaystyle\int\limits_{3}^{19}\color{red}{(-8+6t)}}(3t-4t^2+t3)\)

  57. SolomonZelman
    • one year ago
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    no, not really... you mixed it up a bit.

  58. Abhisar
    • one year ago
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    oh .. !, then?

  59. arindameducationusc
    • one year ago
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    its without -8+6t

  60. arindameducationusc
    • one year ago
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    the same expression you wrote except that!

  61. Abhisar
    • one year ago
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    \[\int\limits_{3}^{13}(3t²-8t+3).du\]

  62. arindameducationusc
    • one year ago
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    right @SolomonZelman ? no 3 to 19 !! @Abhisar

  63. Abhisar
    • one year ago
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    \[\int\limits_{3}^{19}(3t²-8t+3).du\] Sorry , Typo

  64. arindameducationusc
    • one year ago
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    yup, hope so

  65. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle\int\limits_{0}^{4}(-8+6t)(3t^2-8t+3)dt}\) You are starting from \(\color{blue}{u=3t^2-8t+3}\) (You are sort of doing a chain rule for u) \(\large\color{black}{\displaystyle\frac{du}{dt}=6t-8}~~~~\Longrightarrow\) \(\large\color{black}{\displaystyle \color{blue}{du= (-8+6t)dt}}\) \(\large\color{black}{\displaystyle x=0~~~\rightarrow~~~u=3(0)^2-8(0)+3~~~\rightarrow~~~u=3}\) \(\large\color{black}{\displaystyle x=0~~~\rightarrow~~~u=3(4)^2-8(4)+3~~~\rightarrow~~~u=19}\) Now you put in the following into the integal: 1) "u" instead of "3t²-8t+3" 2) "du" instead of "(-8+6t)•dt" 3) Lower limit of u=3 instead of "t=0" 4) Upper limit of u=19 instead of "t=4" YOU GET: \(\large\color{red}{\displaystyle\int\limits_{3}^{19}(u)du}\)

  66. arindameducationusc
    • one year ago
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    nice concept @SolomonZelman I totally got it! WOW!

  67. SolomonZelman
    • one year ago
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    Note that "-8+6t" and "dx" are going away, being replaced by "du".

  68. Abhisar
    • one year ago
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    Ok, so what will be the final answer O.O

  69. SolomonZelman
    • one year ago
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    well what is the integral of u (or, u¹) ?

  70. arindameducationusc
    • one year ago
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    replace t by u, dt by du and put 3 and 19 for the limit

  71. SolomonZelman
    • one year ago
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    \(\LARGE \rm \color{black}{\displaystyle\int\limits_{3}^{19}(u)du=\left.\frac{u^{1\color{royalblue}{+1}}}{1\color{royalblue}{+1}}~\right|}^{_{19}}_{^3}\)

  72. SolomonZelman
    • one year ago
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    arinda what you said is just make everything the same but instead of using "u", use "t" it is going to be the same exact thing, but there is no need in doing that

  73. arindameducationusc
    • one year ago
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    okay @SolomonZelman

  74. Abhisar
    • one year ago
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    oh.....ok, !!!!! u=176. so now \(\sf 3t^2−8t+3=176\) right?

  75. SolomonZelman
    • one year ago
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    why are you doing that, Abhisar?

  76. SolomonZelman
    • one year ago
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    oh the result for the integral is 176 you mean?

  77. Abhisar
    • one year ago
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    Ooops..my apologies, yeah...

  78. SolomonZelman
    • one year ago
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    yes, and just to clarify, we are NOT substituting back the t, once we found the equivalent limits of integration for u and placed THEM in.

  79. Abhisar
    • one year ago
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    ok, one final question. In what cases we can use this method?

  80. SolomonZelman
    • one year ago
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    In many cases. I really want to through a lot of examples rather than giving a verbal definition.

  81. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \int{}\sin^{80}(x)\cos(x)~dx }\) (you can see that u=sin(x), and du=cos(x) dx) AND NOTE: when you find the integral of u, yuo DO need to sub back the x into the result. from now i will post just the type of problems, not their solutions (but if you will be unable to do one of them, then go ahead and ask in this or if you like - a different post) \(\large\color{black}{ \displaystyle \int{}8x^3\sqrt{x^4+5} ~dx }\) \(\large\color{black}{ \displaystyle \int{}x\cos(x^2)~dx}\) (I am using x, but it can be any other letter, I am just used to x) \(\large\color{black}{ \displaystyle \int{}x(x^4-1)~dx}\) (this is a little more advanced tho) want more?

  82. SolomonZelman
    • one year ago
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    mostly, the once that have the derivative of u sitting in the integral.

  83. SolomonZelman
    • one year ago
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    ones*

  84. Abhisar
    • one year ago
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    Let me do them first, I think they will be enough. But thank you very much for the help. I really appreciate it, you have been tutoring me since an hour, that means a lot c:

  85. SolomonZelman
    • one year ago
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    You are always welcome!

  86. Abhisar
    • one year ago
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    I will post my answers in a new thread since this one is becoming annoying to handle c:

  87. SolomonZelman
    • one year ago
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    one note: By indefinite integrals, you always need to sub in the original variable back (after you integrate with the substituted variable).

  88. SolomonZelman
    • one year ago
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    alright, if you like to use a new thread:) Whatever you like!

  89. SolomonZelman
    • one year ago
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    c(u) then

  90. Abhisar
    • one year ago
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    Yeah c; and thanks a bunch again c:

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