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Diyadiya

  • 4 years ago

Evaluate

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  1. wasiqss
    • 4 years ago
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    evaluate diya LOl

  2. Diyadiya
    • 4 years ago
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    Equation editor is not working ?

  3. wasiqss
    • 4 years ago
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    it is !

  4. Diyadiya
    • 4 years ago
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    Okay integrate -1 to 3/2 |x sin(pi x)| dx

  5. wasiqss
    • 4 years ago
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    ok i will

  6. Diyadiya
    • 4 years ago
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    tell me how to start

  7. Diyadiya
    • 4 years ago
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    Equation editor is not working :/

  8. wasiqss
    • 4 years ago
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    ok diya do it by parts i did it

  9. Diyadiya
    • 4 years ago
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    but there is modulus

  10. wasiqss
    • 4 years ago
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    ohh dint c the modulus

  11. wasiqss
    • 4 years ago
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    w8 tellling u

  12. Diyadiya
    • 4 years ago
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    okay

  13. Diyadiya
    • 4 years ago
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    http://ncertbooks.prashanthellina.com/class_12.Mathematics.MathematicsPartII/Integral%20ch_7%2006.11.06.pdf Page number 351 (its written on the top right corner)

  14. wasiqss
    • 4 years ago
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    plz show us how to remove modulus

  15. Diyadiya
    • 4 years ago
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    full solution is there lol

  16. Diyadiya
    • 4 years ago
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    i just wanna know that starting part

  17. razor99
    • 4 years ago
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    ummm guyzzz

  18. Mertsj
    • 4 years ago
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    f=x, f'=dx g'=sin pix g=-1/picospix

  19. Diyadiya
    • 4 years ago
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    Hm What ? sorry can you explain

  20. Mertsj
    • 4 years ago
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    Do you use u and v or do you use f(x) and g(x) when you integrate by parts.?

  21. Diyadiya
    • 4 years ago
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    for parts f(x) and g(x)

  22. Mertsj
    • 4 years ago
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    \[f(x)=x\] \[f'(x)=dx\] \[g'(x)=\sin \pi x\] \[g(x)=-\frac{1}{\pi}\cos \pi \]

  23. wasiqss
    • 4 years ago
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    bt man how to remove modulus

  24. Diyadiya
    • 4 years ago
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    Equation editor isn't working

  25. Diyadiya
    • 4 years ago
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    Can you see the post @wasiqss

  26. Mertsj
    • 4 years ago
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    I thought you wanted to integrate the function. Didn't know you were trying to remove the modulus.

  27. Diyadiya
    • 4 years ago
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    Mertsj i cant see what you posted ,it just shows [maths processing error ]

  28. wasiqss
    • 4 years ago
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    diya i can see tho :P

  29. Diyadiya
    • 4 years ago
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    Wait let me refresh

  30. wasiqss
    • 4 years ago
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    mertsj jus remove the modulus first then i will integrate

  31. Mertsj
    • 4 years ago
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    |dw:1330277056196:dw|

  32. Mertsj
    • 4 years ago
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    Is that the problem or not?

  33. wasiqss
    • 4 years ago
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    yes

  34. Mertsj
    • 4 years ago
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    Then integrate it using the substitutions I gave you.

  35. Diyadiya
    • 4 years ago
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    |dw:1330277124894:dw|

  36. wasiqss
    • 4 years ago
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    modulus is thea man

  37. Diyadiya
    • 4 years ago
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  38. Mertsj
    • 4 years ago
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    f(x)=x f'(x)=dx g'(x)=sin pi x g(x)= -1/pi cos pi x

  39. Diyadiya
    • 4 years ago
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    So we dont have to remove modulus first ?

  40. wasiqss
    • 4 years ago
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    yeh diya thats my question

  41. Mertsj
    • 4 years ago
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    Why would you? Integrate the function.

  42. wasiqss
    • 4 years ago
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    mertsj we have do something with modulus first

  43. Mertsj
    • 4 years ago
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    Why?

  44. Diyadiya
    • 4 years ago
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    This is the solution given in my book

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  45. wasiqss
    • 4 years ago
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    diya this is techniqu for removing modulus that u need to learn from someone cox im not able to recall it

  46. Mertsj
    • 4 years ago
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    Right. Since it's absolute value there are actually two integrals to do.

  47. Mertsj
    • 4 years ago
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    Where was the modulus removed in diyadiya's example post?

  48. Mertsj
    • 4 years ago
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    Which step?

  49. Zarkon
    • 4 years ago
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    find where x sin(pi x) is positive and negative

  50. wasiqss
    • 4 years ago
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    as in how

  51. Mr.Math
    • 4 years ago
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    Which part of the solution don't you understand Diya?

  52. Diyadiya
    • 4 years ago
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    the beginning

  53. Mr.Math
    • 4 years ago
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    Alright. You must know that the definition of absolute value, that's |x|=x, if x>=0 and -x if x<0. So we have to find when \(x\sin(\pi x)\) is positive and when it's negative.

  54. Diyadiya
    • 4 years ago
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    MrMath i cant see Latex

  55. Mr.Math
    • 4 years ago
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    Lets first consider the interval from (-1,0). x is obviously negative in this interval, and so is sin(pi*x) because sin(pi x) would lie either in the third or fourth quadrant for x in (-1,0). And hence xsin(pi*x) is the product of two negative number which is positve in (-1,0). Following so far?

  56. Mr.Math
    • 4 years ago
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    Alright. You must know that the definition of absolute value, that's |x|=x, if x>=0 and -x if x<0. So we have to find when xsin(πx) is positive and when it's negative.

  57. Diyadiya
    • 4 years ago
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    Why is it in the 3rd or 4th Quadrant ?

  58. Mr.Math
    • 4 years ago
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    Because when x in (-1,0), the angle, πx in (-π,0), right?

  59. Diyadiya
    • 4 years ago
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    yeah

  60. Mr.Math
    • 4 years ago
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    So you know why sin(πx) would be negative in this interval?

  61. Diyadiya
    • 4 years ago
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    okay sine is negetive in 3rd & 4th Quadrant

  62. Mr.Math
    • 4 years ago
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    And then the xsin(πx) is positve in (-1,0).

  63. Mr.Math
    • 4 years ago
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    Now in the interval (0,1) is xsin(πx) positive or negative?

  64. Diyadiya
    • 4 years ago
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    negative ?

  65. Mr.Math
    • 4 years ago
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    Lets cut it into two pieces. Is x positive or negative in (0,1)?

  66. Diyadiya
    • 4 years ago
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    wait a minute , i'm trying to understand the previous thing :D 1min

  67. Mr.Math
    • 4 years ago
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    Take your time :-)

  68. Diyadiya
    • 4 years ago
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    Alright

  69. Diyadiya
    • 4 years ago
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    x is positive in (0,1)

  70. Mr.Math
    • 4 years ago
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    Good. What about sin(πx)?

  71. Diyadiya
    • 4 years ago
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    Positive

  72. Mr.Math
    • 4 years ago
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    Great. So xsin(πx) over all is obviously..?

  73. Diyadiya
    • 4 years ago
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    positive

  74. Mr.Math
    • 4 years ago
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    What is the next interval we should consider now?

  75. Diyadiya
    • 4 years ago
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    (1,3/2) ?

  76. Mr.Math
    • 4 years ago
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    Right, (1,3/2).

  77. Mr.Math
    • 4 years ago
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    And xsin(πx) will be positive or negative in this interval?

  78. Diyadiya
    • 4 years ago
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    positive

  79. Mr.Math
    • 4 years ago
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    Why?

  80. Diyadiya
    • 4 years ago
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    hm x is positive in (1,3/2)

  81. Diyadiya
    • 4 years ago
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    its not ?

  82. Mr.Math
    • 4 years ago
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    Correct. But what about sin(πx)?

  83. Diyadiya
    • 4 years ago
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    positive ?

  84. Diyadiya
    • 4 years ago
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    wait

  85. Diyadiya
    • 4 years ago
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    No its negative

  86. Diyadiya
    • 4 years ago
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    since the angle is b/w (pi ,3pi/2)

  87. Mr.Math
    • 4 years ago
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    Very good. That means that xsin(πx) will be negative in (1,3/2), right?

  88. Diyadiya
    • 4 years ago
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    Yeah

  89. Mr.Math
    • 4 years ago
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    So now since you know where our expression is positive and negative you can rewrite it using the definition of absolute value. Can you do that?

  90. Mr.Math
    • 4 years ago
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    We have found that it's positive in (-1,1) and negative in (1,3/2). You agree?

  91. Diyadiya
    • 4 years ago
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    why (-1,1) ?

  92. Mr.Math
    • 4 years ago
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    Didn't we just find that it's positive in (-1,0) and in (0,1)?

  93. Diyadiya
    • 4 years ago
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    Oh yes !!

  94. Mr.Math
    • 4 years ago
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    Okay, the next step is to write it as it's written in your book. :-)

  95. Diyadiya
    • 4 years ago
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    Yeah Got it :D

  96. Diyadiya
    • 4 years ago
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    Thank You soo much :D that was a great explaination :D

  97. Mr.Math
    • 4 years ago
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    Awesome!

  98. Diyadiya
    • 4 years ago
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    Now i can do the rest :-) Thanks again :)

  99. Mr.Math
    • 4 years ago
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    You're welcome!

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