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hiralpatel121 Group Title

How can I find the indefinite integral of the attached problem?

  • 2 years ago
  • 2 years ago

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  1. hiralpatel121 Group Title
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    • 2 years ago
  2. Spacelimbus Group Title
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    a substitution will do, let =3x

    • 2 years ago
  3. Spacelimbus Group Title
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    let u=3x*

    • 2 years ago
  4. hiralpatel121 Group Title
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    When I solved it, I got:

    • 2 years ago
  5. Spacelimbus Group Title
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    \[ u = 3x \longrightarrow \frac{du}{dx}=3 \\ \frac{2}{3}\cdot \frac{du}{dx}=2 \] \[ \large \frac{2}{3}\int u^{-\frac{1}{3}}du\]

    • 2 years ago
  6. Spacelimbus Group Title
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    can you go from here?

    • 2 years ago
  7. hiralpatel121 Group Title
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    I don't know what to do after this..

    • 2 years ago
  8. Spacelimbus Group Title
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    you integrate and backsubstitute. you will get: \[ \sqrt[3]{u}^2\]

    • 2 years ago
  9. hiralpatel121 Group Title
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    So that's the answer?

    • 2 years ago
  10. Spacelimbus Group Title
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    the answer is \[ \sqrt[3]{3x}^2 \]

    • 2 years ago
  11. hiralpatel121 Group Title
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    What do you do after you integrate and back substitute? I want to show all the steps.

    • 2 years ago
  12. Spacelimbus Group Title
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    Then you are done. That's the indefinite integral.

    • 2 years ago
  13. hiralpatel121 Group Title
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    But the answer is cube root of 3x^2. You did something after you got cube root of u^2.

    • 2 years ago
  14. Spacelimbus Group Title
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    well at the beginning I said substitute u=3x, and you want your result to be in forms of x right? so you have to back-substitute, where there is an u, you put a 3x instead.

    • 2 years ago
  15. hiralpatel121 Group Title
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    Oh okay, can you help me with one more?

    • 2 years ago
  16. Spacelimbus Group Title
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    sure

    • 2 years ago
  17. hiralpatel121 Group Title
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    • 2 years ago
  18. Spacelimbus Group Title
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    Do you know basic integration of exponents yes?

    • 2 years ago
  19. hiralpatel121 Group Title
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    Some of it.

    • 2 years ago
  20. Spacelimbus Group Title
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    because you can write this integral like this: \[ \int (x^{-2} -x^{-3})dx\]

    • 2 years ago
  21. Spacelimbus Group Title
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    And then you just integrate normally, increase the exponent by one and divide by the new exponent.

    • 2 years ago
  22. hiralpatel121 Group Title
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    Can we do it step by step?

    • 2 years ago
  23. Spacelimbus Group Title
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    yes one second

    • 2 years ago
  24. hiralpatel121 Group Title
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    Okay

    • 2 years ago
  25. Spacelimbus Group Title
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    so you can actually split this integral up if that makes it easier for you \[ \int x^{-2} dx- \int x^{-3}dx \]

    • 2 years ago
  26. Spacelimbus Group Title
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    if you integrate the first term you should get \[- \frac{1}{x}\]

    • 2 years ago
  27. hiralpatel121 Group Title
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    Okay, I got that part.

    • 2 years ago
  28. Spacelimbus Group Title
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    think you can figure out the second one? it works exactly the same.

    • 2 years ago
  29. hiralpatel121 Group Title
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    Wait, I mean -1/2x^2

    • 2 years ago
  30. Spacelimbus Group Title
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    yes you did it right

    • 2 years ago
  31. hiralpatel121 Group Title
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    It's the first or second one?

    • 2 years ago
  32. Spacelimbus Group Title
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    \[ \frac{x^{-3+1}}{-3+1} = -\frac{1}{2x^2}\]

    • 2 years ago
  33. hiralpatel121 Group Title
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    Okay, what is the step after that?

    • 2 years ago
  34. Spacelimbus Group Title
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    That's it, you have taken the indefinite integral so you can just write it down. usually you add a + C at the end for indefinite integrals to show that an arbitrary constant could have been lost.

    • 2 years ago
  35. hiralpatel121 Group Title
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    So it would be: -1/2x^2+C?

    • 2 years ago
  36. Spacelimbus Group Title
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    no, both of them, you have taken two indefinite integrals remember? Look at the formula I have posted above, we did split them up, you have derived the first one and the second one, both are part of your solution.

    • 2 years ago
  37. hiralpatel121 Group Title
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    -1/x - 1/2x^2 + C = Answer?

    • 2 years ago
  38. Spacelimbus Group Title
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    \[ \int x^{-2} dx- \int x^{-3}dx = -\frac{1}{x} + \frac{1}{x^2}+C \]

    • 2 years ago
  39. Spacelimbus Group Title
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    almost, the minus becomes a plus, because there is a minus in front of the integral. See it?

    • 2 years ago
  40. hiralpatel121 Group Title
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    Yes, but wasn't the second part -1/ 2x^2? Then in the answer, why did you leave out the 2 in front of the x^2?

    • 2 years ago
  41. Spacelimbus Group Title
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    oops, you are absolutely right, I messed it up

    • 2 years ago
  42. Spacelimbus Group Title
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    \[ \int x^{-2} dx- \int x^{-3}dx = -\frac{1}{x} + \frac{1}{2x^2}+C \]

    • 2 years ago
  43. Spacelimbus Group Title
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    well done! (-: you discovered my mistake.

    • 2 years ago
  44. hiralpatel121 Group Title
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    Yay! I need help with a few more if you have time?

    • 2 years ago
  45. Spacelimbus Group Title
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    I believe so. just post them.

    • 2 years ago
  46. hiralpatel121 Group Title
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    • 2 years ago
  47. Spacelimbus Group Title
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    you know how to setup this integral?

    • 2 years ago
  48. hiralpatel121 Group Title
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    No

    • 2 years ago
  49. hiralpatel121 Group Title
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    I'm not good with trigs..at all :/

    • 2 years ago
  50. Spacelimbus Group Title
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    hmm well I more meant, do you know what you have to do in order to calculate the area of this ?

    • 2 years ago
  51. hiralpatel121 Group Title
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    No

    • 2 years ago
  52. Spacelimbus Group Title
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    The integral calculates the Area below a graph, do you know that?

    • 2 years ago
  53. Spacelimbus Group Title
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    So whenever you want to calculate the area underneath a graph of a function, of any kind, you want to setup an integral.

    • 2 years ago
  54. Spacelimbus Group Title
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    \[ \large \int_a^b f(x) dx \] this is the general setup of an integral, it is how it looks like, y=f(x) is the function you want to integrate, and the dx just tells you in respect to what you have to integrate it with.

    • 2 years ago
  55. Spacelimbus Group Title
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    a and b are the boundaries of your integral, more correctly it should be written like this \[ \int_{x=a}^{x=b} f(x)dx\]

    • 2 years ago
  56. Spacelimbus Group Title
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    for a < b, that means you integrate a function, starting from the point a, and you integrate it's entire area until you hit the point b.

    • 2 years ago
  57. Spacelimbus Group Title
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    the under bound is always the lower limits, meaning lower value, the upper bound is always the upper limit, meaning higher value.

    • 2 years ago
  58. hiralpatel121 Group Title
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    Okay

    • 2 years ago
  59. Spacelimbus Group Title
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    So do you have the picture open?

    • 2 years ago
  60. hiralpatel121 Group Title
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    Yes, I have it open

    • 2 years ago
  61. Spacelimbus Group Title
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    So can you tell me what the boundaries (upper and lower) for this integral are?, from where to where does it run? Look at the area you want to integrate.

    • 2 years ago
  62. hiralpatel121 Group Title
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    0 to pi/2

    • 2 years ago
  63. Spacelimbus Group Title
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    exactly!

    • 2 years ago
  64. Spacelimbus Group Title
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    and the function is y=x+cosx right?

    • 2 years ago
  65. hiralpatel121 Group Title
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    Yes

    • 2 years ago
  66. Spacelimbus Group Title
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    so this means it looks like that : \[ \large \int_0^{\frac{\pi}{2}}(x + \cos x)dx =\int_0^{\frac{\pi}{2}}xdx + \int_0^{\frac{\pi}{2}}\cos(x)dx \]

    • 2 years ago
  67. Spacelimbus Group Title
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    Both are easy integrals for themselves, when you look at them individually, keep in mind, the integrals turns a function into it's antiderivative, that means if you derive an integrated function, you should get back to the original function.

    • 2 years ago
  68. hiralpatel121 Group Title
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    Okay, that's all there is to it?

    • 2 years ago
  69. Spacelimbus Group Title
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    well now you have to evaluate it, you want to determine the area of the given region, that means you want to compute the integral and evaluate it.

    • 2 years ago
  70. Spacelimbus Group Title
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    just like you did before, take the antiderivative, only this time you have values to plug in and see what happens.

    • 2 years ago
  71. hiralpatel121 Group Title
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    Sorry, I'm so confused. This is due by 9:30 and I'm under stress. Ugh!

    • 2 years ago
  72. Spacelimbus Group Title
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    ohhh, I see.

    • 2 years ago
  73. hiralpatel121 Group Title
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    So, now I plug in pi/2 and then plug in 0?

    • 2 years ago
  74. Spacelimbus Group Title
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    first you need to compute the antiderivate.

    • 2 years ago
  75. Spacelimbus Group Title
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    have you done that? can you tell me what the integral of x is ?

    • 2 years ago
  76. hiralpatel121 Group Title
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    x^2/2

    • 2 years ago
  77. Spacelimbus Group Title
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    exactly!

    • 2 years ago
  78. Spacelimbus Group Title
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    and see, there you plug in the values now, it always works like this, plug in the higher value, then subtract from that the value that you get when you plug in the lower value.

    • 2 years ago
  79. Spacelimbus Group Title
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    Usually you see this written like that \[ \left. \frac{x^2}{2} \right|_{x=0}^{x=\frac{\pi}{2}}\ = \frac{\frac{\pi ^2}{4}}{2} - \frac{0^2}{2} = \frac{\pi^2}{8} \]

    • 2 years ago
  80. Spacelimbus Group Title
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    but that's only the first step, now you need to do the exact same with the cos(x) function, your second integral.

    • 2 years ago
  81. hiralpatel121 Group Title
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    That's the difficult part. Wouldn't I need a unit circle?

    • 2 years ago
  82. Spacelimbus Group Title
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    no, you need to know the trig functions and their derivative by heart the derivative of sin(x) = cos(x) therefore the antiderivative of cos(x) = sin(x)

    • 2 years ago
  83. hiralpatel121 Group Title
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    Wouldn't it be negative sin(x)?

    • 2 years ago
  84. Spacelimbus Group Title
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    no the derivative of cos(x) is -sin(x)

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  85. Spacelimbus Group Title
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    but we are doing integration , that's the other way around.

    • 2 years ago
  86. hiralpatel121 Group Title
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    Okay, now I just plug in pi/2 into sin x?

    • 2 years ago
  87. Spacelimbus Group Title
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    exactly and then you subtract from that sin(0) , but sin(0) =0, so you don't have to worry about that, just to empathize the order.

    • 2 years ago
  88. hiralpatel121 Group Title
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    I get sin(pi/2) = 1

    • 2 years ago
  89. Spacelimbus Group Title
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    that's true!

    • 2 years ago
  90. Spacelimbus Group Title
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    so the entire area is?

    • 2 years ago
  91. hiralpatel121 Group Title
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    -1 ?

    • 2 years ago
  92. Spacelimbus Group Title
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    no you add them both together now, just like we did before with the indefinite integrals, now you add the values together

    • 2 years ago
  93. hiralpatel121 Group Title
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    So it's 1.

    • 2 years ago
  94. Spacelimbus Group Title
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    yes, but that's not the entire area

    • 2 years ago
  95. Spacelimbus Group Title
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    Remember what we got for the first integral?

    • 2 years ago
  96. Spacelimbus Group Title
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    \[\large \int_0^{\frac{\pi}{2}}(x + \cos x)dx =\int_0^{\frac{\pi}{2}}xdx + \int_0^{\frac{\pi}{2}}\cos(x)dx \\ = \underline{\underline{\frac{\pi^2}{8}+1}}\]

    • 2 years ago
  97. hiralpatel121 Group Title
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    Now I have to subtract pi/8+1 by 1?

    • 2 years ago
  98. Spacelimbus Group Title
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    no, that's the solution already up there, I did underline it.

    • 2 years ago
  99. hiralpatel121 Group Title
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    Ohhh so the entire area is pi/8 +1.

    • 2 years ago
  100. Spacelimbus Group Title
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    see \(\frac{\pi^2}{8}\) is what we got for the first integral, when we evaluated it, remember? x^2/2, you did that yourself, now the next integral - it has a plus there, you evaluated that integral and you said the area is 1 so the entire area is both summed together.

    • 2 years ago
  101. hiralpatel121 Group Title
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    Yes so how would I write the final answer?

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  102. Spacelimbus Group Title
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    just like I did, up there

    • 2 years ago
  103. Spacelimbus Group Title
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    \[ A = \underline{\underline{\frac{\pi^2}{8}+1}}\]

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  104. Spacelimbus Group Title
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    because that is the area.

    • 2 years ago
  105. hiralpatel121 Group Title
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    Okay, thank you so much! Are you good with true and false questions? I have the answers already but I have to write sentences and I need help with that.

    • 2 years ago
  106. Spacelimbus Group Title
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    we can give it a try.

    • 2 years ago
  107. hiralpatel121 Group Title
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    • 2 years ago
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  108. Spacelimbus Group Title
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    one you answered right

    • 2 years ago
  109. hiralpatel121 Group Title
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    How do I explain it in sentences though?

    • 2 years ago
  110. Spacelimbus Group Title
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    two you answered right two, but your reasoning isn't quite correct

    • 2 years ago
  111. Spacelimbus Group Title
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    False: The integral of two functions cannot be computed by taking the integral of each function individually and multiplying the results. The correct method is using partial integration.

    • 2 years ago
  112. hiralpatel121 Group Title
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    If I use partial integration, how would it look?

    • 2 years ago
  113. Spacelimbus Group Title
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    \[ \large\int f'(x)g(x)dx= f(x)g(x)- \int f(x)g'(x) dx \]

    • 2 years ago
  114. hiralpatel121 Group Title
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    Okay

    • 2 years ago
  115. hiralpatel121 Group Title
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    For 3-5, I just don't know how to word it.

    • 2 years ago
  116. Spacelimbus Group Title
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    false because y= ln e =1 is a number, the derivative of a number is always zero.

    • 2 years ago
  117. hiralpatel121 Group Title
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    That's all I have to say for that?

    • 2 years ago
  118. Spacelimbus Group Title
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    totally

    • 2 years ago
  119. Spacelimbus Group Title
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    for 3.

    • 2 years ago
  120. hiralpatel121 Group Title
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    Okay, great, 4 and 5 left.

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  121. Spacelimbus Group Title
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    4 and 5 you answered perfectly already.

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  122. Spacelimbus Group Title
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    there is nothing I would add.

    • 2 years ago
  123. hiralpatel121 Group Title
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    But I have to rephrase it all into sentence form and I don't know how.

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  124. Spacelimbus Group Title
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    ah yeh? Well, this is math, sometimes showing the example is the best way to make a phrase.

    • 2 years ago
  125. Spacelimbus Group Title
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    You can always make sentences though of course, for 4 you can write. False, if you derive the natural log of x you get the inverse function, when you integrate it you again have to use the technic of partial integration which leads to xlnx-x + C

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  126. Spacelimbus Group Title
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    For 5. There is a function which is equal to it's derivative, this function is e^x, it's slope at any given point of x is equal to it's function valued at given point.

    • 2 years ago
  127. hiralpatel121 Group Title
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    Okay, thank you so so much!!

    • 2 years ago
  128. Spacelimbus Group Title
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    welcome

    • 2 years ago
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