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anonymous

  • one year ago

intergration problem

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    my firsr step is antderv right

  3. anonymous
    • one year ago
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    i have an idea (although i totally suck at these)

  4. anonymous
    • one year ago
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    it will be easier to integrate in terms of \(y\) not \(x\) so lets solve each of these for \(x\)

  5. anonymous
    • one year ago
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    the first one is \(x=y+1\) and the second is \(x=\frac{y^2-6}{2}\)

  6. anonymous
    • one year ago
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    we also need to find the limits of integration did you find where they intersect?

  7. anonymous
    • one year ago
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    oh and to answer your question, no, the first step is not to find the anti derivative, the first step is to find the thing you need to integrate, both the integrand and the limits of integration THEN you can take the anti - derivative to compute the integral

  8. anonymous
    • one year ago
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    how to do that

  9. anonymous
    • one year ago
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    set them equal and solve

  10. anonymous
    • one year ago
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    \[y=x-1,y^2=2x+6\] replace the \(y\) in the second equation by \(x-1\) and solve \[(x-1)^2=2x+6\] turns out you get an easy quadratic equation to solve (it factor, get integer solutions )

  11. anonymous
    • one year ago
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    that is if you want to do it by hand it if was me, i would cut to the chase http://www.wolframalpha.com/input/?i=%28x-1%29^2%3D2x%2B6

  12. anonymous
    • one year ago
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    k so i get x^2-4x+8

  13. anonymous
    • one year ago
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    you get \(x=-1\) or \(x=5\) the points where they intersect are \((-1,-2)\) and \((5,4)\)

  14. anonymous
    • one year ago
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    hmm no i think if you do it by hand you get \[x^2-2x+1=2x+6\\ x^2-4x-5=0\]

  15. anonymous
    • one year ago
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    o wow yea yea

  16. anonymous
    • one year ago
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    factors as\[(x-5)(x+1)=0\\ x=-1,x=5\]

  17. anonymous
    • one year ago
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    ok then

  18. anonymous
    • one year ago
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    now this is a parabola that opens sideways so it will be easier to integrate wrt y not x

  19. anonymous
    • one year ago
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    here is a picture http://www.wolframalpha.com/input/?i=+y%3Dx+-+1%2C+y^2%3D2x%2B6

  20. anonymous
    • one year ago
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    so if you want to integrate wrt y, you need to write each of these as a function of \(y\) i.e solve each for \(x\)

  21. anonymous
    • one year ago
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    first one is \[x=y+1\] second is \[x=\frac{y^2-6}{2}\]

  22. anonymous
    • one year ago
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    so far so good?

  23. anonymous
    • one year ago
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    okey

  24. anonymous
    • one year ago
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    now if we turn it sideways we see the line is above the parabola the limits of integration in terms of y means we use the y values \[\int_{-2}^5y+1-(\frac{y^2-6}{2})dy\]

  25. anonymous
    • one year ago
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    do some algebra first maybe, then compute that integral

  26. anonymous
    • one year ago
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    i have to take the antiderv right

  27. anonymous
    • one year ago
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    i would do the algebra first then take anti derivatives

  28. anonymous
    • one year ago
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    oh u mean simplify

  29. anonymous
    • one year ago
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    oops \[\int_{-2}^5\left(\frac{-y^2}{2}+y+4\right)dy\]

  30. xapproachesinfinity
    • one year ago
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    seems good :)

  31. anonymous
    • one year ago
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    thanks !

  32. anonymous
    • one year ago
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    now take the anto right

  33. anonymous
    • one year ago
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    NOW take anti derivatives

  34. anonymous
    • one year ago
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    yeah i know you are dying to do it, do it now

  35. anonymous
    • one year ago
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    okey tge y2/2 is confusing

  36. anonymous
    • one year ago
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    would it be y3/6?

  37. anonymous
    • one year ago
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    no

  38. xapproachesinfinity
    • one year ago
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    yes! just don't forgot - sign

  39. anonymous
    • one year ago
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    \[-\frac{y^2}{6}\] for the first term

  40. anonymous
    • one year ago
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    ok as @xapproachesinfinity said "yes" only "no"

  41. anonymous
    • one year ago
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    oh yea sorry

  42. anonymous
    • one year ago
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    the rest should be routine anti derivative is \[-\frac{y^2}{6}+\frac{y^2}{2}+4y\] plug in 5, plug in -2 and subtract if you do it right, you will get 18i think

  43. anonymous
    • one year ago
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    so i get 12.5 after simliy

  44. anonymous
    • one year ago
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    damn typo \[-\frac{y^3}{6}+\frac{y^2}{2}+4y\]

  45. anonymous
    • one year ago
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    i think ur right one my firends got 18 two

  46. anonymous
    • one year ago
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    me, i like to know what the answer is before i begin http://www.wolframalpha.com/input/?i=+area+between+the+curves+y%3Dx+-+1%2C+y^2%3D2x%2B6

  47. anonymous
    • one year ago
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    ok so i have more intergral problems but just much eaiser i am just stuck since i am new to all this so could u help me more?

  48. anonymous
    • one year ago
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    ok sure why not

  49. anonymous
    • one year ago
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    if i can i kind of suck at these though

  50. anonymous
    • one year ago
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    ok give me 5 minutes i am finishing somthing

  51. anonymous
    • one year ago
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    kk

  52. anonymous
    • one year ago
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    lets see

  53. anonymous
    • one year ago
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    |dw:1433296429808:dw|

  54. anonymous
    • one year ago
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    i did everything but getting the answer wrong

  55. anonymous
    • one year ago
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    here is my work just before the solution

  56. anonymous
    • one year ago
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    2/3x^3/2 + x +lnx , am i right

  57. anonymous
    • one year ago
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    looks like you have an extra term there

  58. anonymous
    • one year ago
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    \[\frac{\sqrt{x}+1}{x}=x^{-\frac{1}{2}}+\frac{1}{x}\]

  59. anonymous
    • one year ago
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    not sure where the x came from

  60. anonymous
    • one year ago
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    how is squrtx , x ^-1/2? half should be positve right

  61. anonymous
    • one year ago
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    also note that \[\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}\] not \(\sqrt{x}\)

  62. anonymous
    • one year ago
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    or if you like \[\frac{\sqrt{x}}{x}=\frac{x^{\frac{1}{2}}}{x}=x^{\frac{1}{2}-1}=x^{-\frac{1}{2}}\]

  63. anonymous
    • one year ago
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    you got that?

  64. anonymous
    • one year ago
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    yes

  65. anonymous
    • one year ago
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    wait i thiught it was +1 when taking anti

  66. anonymous
    • one year ago
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    ok now take the anti derivative

  67. anonymous
    • one year ago
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    yes we didn't do that yet

  68. anonymous
    • one year ago
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    then why did u -1?

  69. anonymous
    • one year ago
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    as you put it, we 'simplified' first to get \[x^{-\frac{1}{2}}+\frac{1}{x}\] now go from there

  70. anonymous
    • one year ago
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    ok lets go slow

  71. anonymous
    • one year ago
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    first off you are given \[\frac{\sqrt{x}+1}{x}\] right?

  72. anonymous
    • one year ago
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    yes

  73. anonymous
    • one year ago
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    then BEFORE we find the anti derivative we are going to divide each term in the numerator by x that is not integrating, that is dividing

  74. anonymous
    • one year ago
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    \[\frac{\sqrt{x}+1}{x}=\frac{\sqrt{x}}{x}+\frac{1}{x}\] right?

  75. anonymous
    • one year ago
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    yes

  76. anonymous
    • one year ago
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    we still have not taken the anti derivative yet the anti derivative of \(\frac{1}{x}\) is what you said \(\ln(x)\)

  77. anonymous
    • one year ago
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    but in order to take the anti derivative of \[\frac{\sqrt{x}}{x}\] we need to write it as x to some power

  78. anonymous
    • one year ago
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    ohh k

  79. anonymous
    • one year ago
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    then we can use the power rule (backwards) for finding it as a power it is \[\frac{\sqrt{x}}{x}=x^{-\frac{1}{2}}\] that is not finding the anti derivative, that is just writing it in exponential form

  80. anonymous
    • one year ago
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    yea i think that my only mistake

  81. anonymous
    • one year ago
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    THEN you can add one to the exponent etc

  82. anonymous
    • one year ago
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    you should get \[2x^{\frac{1}{2}}\] or if you prefer \[2\sqrt{x}\]

  83. anonymous
    • one year ago
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    questions or clear?

  84. anonymous
    • one year ago
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    k i am going to finish this uo

  85. anonymous
    • one year ago
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    kk

  86. anonymous
    • one year ago
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    i am getting 6.38 , u ?

  87. anonymous
    • one year ago
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    @satellite73 the answer on my book is like 2+ln4

  88. anonymous
    • one year ago
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    @satellite73 u their ?

  89. anonymous
    • one year ago
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    plug in 4, plug in 1 subtract

  90. anonymous
    • one year ago
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    if the book as \(\ln(4)\) as part of the answer, do not use a calculator you may have got the same answer but as a decimal

  91. anonymous
    • one year ago
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    \[F(x)=2\sqrt{x}+\ln(x)\]

  92. anonymous
    • one year ago
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    \[F(4)=2\sqrt4 +\ln(4)=4+\ln(4)\]

  93. anonymous
    • one year ago
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    \[F(1)=2\sqrt 1+\ln(1)=2\]

  94. anonymous
    • one year ago
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    i did 1^1/2 i got 1

  95. anonymous
    • one year ago
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    did you forget the 2 out front?

  96. anonymous
    • one year ago
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    2-1=1

  97. anonymous
    • one year ago
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    omg yea

  98. anonymous
    • one year ago
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    \[\int x^{-\frac{1}{2}}dx=2\sqrt{x}\]

  99. anonymous
    • one year ago
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    damn silly mistakes errrr

  100. anonymous
    • one year ago
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    lol there are lots of ways to make a mistake!

  101. anonymous
    • one year ago
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    serioulsy the only reasons i lose mrks in calc

  102. anonymous
    • one year ago
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    but i am making miskates here so i wont makw them on the test

  103. anonymous
    • one year ago
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    good!

  104. anonymous
    • one year ago
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    thats the number 2 reason for doing homework

  105. anonymous
    • one year ago
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    do u know questions realted to area under graph where u do the same thing just now draw

  106. anonymous
    • one year ago
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    go ahead i will look

  107. anonymous
    • one year ago
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    "fnid teh aera fo teh reiogn beteewn the gienv curesv, setch of the reiogn

  108. anonymous
    • one year ago
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    y=x^2 +3 and y= x+a from 2 to 4

  109. anonymous
    • one year ago
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    i mean x+1

  110. anonymous
    • one year ago
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    what i did was made them equal to each other then anti

  111. anonymous
    • one year ago
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    no not in this case

  112. anonymous
    • one year ago
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    you are given the limits of integration, you don't have to find them

  113. anonymous
    • one year ago
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    the only thing you need to know is that the parabola lies above the line

  114. anonymous
    • one year ago
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    so \[\int_2^2\left(x^2+3-(x+1)\right)dx\]

  115. anonymous
    • one year ago
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    simplify then anti ?

  116. anonymous
    • one year ago
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    no first algebra, then anti derivative

  117. anonymous
    • one year ago
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    so its x2 -x +2

  118. anonymous
    • one year ago
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    so after antoi i get the wuation 1/3x3 - 1/2x2 +2x , am i right

  119. anonymous
    • one year ago
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    yes looks good

  120. anonymous
    • one year ago
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    you see the difference in that and the previous one is that you were asked for the area between the curves then you had to find where they intersect in this case you are given the limits of integration

  121. anonymous
    • one year ago
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    ok

  122. anonymous
    • one year ago
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    then what do we do

  123. anonymous
    • one year ago
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    plug in 4, plug in 2, subtract

  124. anonymous
    • one year ago
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    that is what you always do \[F(x)=\frac{x^3}{3}-\frac{x^2}{2}+2x\] compute \[F(4)-F(2)\]

  125. anonymous
    • one year ago
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    what do u get , i get 12.05

  126. anonymous
    • one year ago
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    idk i didn't do it

  127. anonymous
    • one year ago
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    ok what do now

  128. anonymous
    • one year ago
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    i will check your answer

  129. anonymous
    • one year ago
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    i got \(\frac{50}{3}\) hmmm

  130. anonymous
    • one year ago
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    oh ur right maybe i did wrong sub

  131. anonymous
    • one year ago
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    \[F(4)=\frac{4^3}{3}-\frac{4^2}{2}+2\times 4\] \[F(2)=\frac{2^3}{3}+\frac{2^2}{4}+2\times 2\]need a calculator or sommat

  132. anonymous
    • one year ago
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    i think i need to stop subbing EVERYRthing into calc

  133. anonymous
    • one year ago
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    ok so what now

  134. anonymous
    • one year ago
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    how can i draw the graph from this info

  135. anonymous
    • one year ago
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    now you are done go have a beer

  136. anonymous
    • one year ago
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    if you have to graph, graph the line and the parabola, that is all

  137. anonymous
    • one year ago
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    both are real easy to graph

  138. anonymous
    • one year ago
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    sooo??|dw:1433301650060:dw|

  139. anonymous
    • one year ago
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    |dw:1433301637176:dw|

  140. anonymous
    • one year ago
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    yeah

  141. anonymous
    • one year ago
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    on the back of my book its like |dw:1433301723404:dw|

  142. anonymous
    • one year ago
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    why is that area shaded

  143. anonymous
    • one year ago
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    because that is what you were asked for the area bounded by the cuve \(y=x^2+3\) the line \(y=x+1\) between the vertical lines \(x=2\) and \(x=4\)

  144. anonymous
    • one year ago
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    so all the work we did had nothing to do with drawing the graph?

  145. anonymous
    • one year ago
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    we found the area of that region, but no, finding the area is not the same as graphing the region, they are two different things

  146. anonymous
    • one year ago
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    how long r u gonna stay here?

  147. anonymous
    • one year ago
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    lol not much longer i hope you got more?

  148. anonymous
    • one year ago
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    i wanna do more probs , i might get stuck again

  149. anonymous
    • one year ago
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    their is a question

  150. anonymous
    • one year ago
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    find the area under the given curve from a to b

  151. anonymous
    • one year ago
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    y=x2 +1 from 0 to 4

  152. anonymous
    • one year ago
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    that is real real easy

  153. anonymous
    • one year ago
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    set up the integral and do it

  154. anonymous
    • one year ago
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    we can graph it even before we solve it right ? and whats the difference

  155. anonymous
    • one year ago
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    why bother, you know what \(x^2+1\) looks like right? all you have to compute here is \[\int_0^4(x^2+1)dx\]

  156. anonymous
    • one year ago
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    idk the answer on the back has a graph

  157. anonymous
    • one year ago
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    that is because... they have a graph in the back the graphics jack up the price of the book

  158. anonymous
    • one year ago
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    probs

  159. anonymous
    • one year ago
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    that integral is real real easy right?

  160. anonymous
    • one year ago
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    yea

  161. anonymous
    • one year ago
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    lots easier than the first two for sure

  162. anonymous
    • one year ago
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    their are different chapters on my homwork but all the same thing

  163. anonymous
    • one year ago
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    one says "area under curve " and the other "area between curves"

  164. anonymous
    • one year ago
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    under the curve means below the curve and above the x axis between the curve means between the curves the first one you just use the limits of integration given you the second you need the upper curve minus the lower curve

  165. anonymous
    • one year ago
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    |dw:1433302834136:dw|

  166. anonymous
    • one year ago
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    didnt we do upper minus bottom in both of them

  167. anonymous
    • one year ago
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    not for the last one no

  168. anonymous
    • one year ago
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    so the last one i asked was area under the curve

  169. anonymous
    • one year ago
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    i toke the anti of x2 + 1

  170. anonymous
    • one year ago
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    i got x3/3 + x

  171. anonymous
    • one year ago
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    then i did F(4)-F(0)

  172. anonymous
    • one year ago
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    yes

  173. anonymous
    • one year ago
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    final soltuion 76/3 which is correct is that top minus bottom ?

  174. anonymous
    • one year ago
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    \[F(0)=0\] in this case it should be easier than most

  175. anonymous
    • one year ago
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    \[\frac{4^3}{3}+4\] is all

  176. anonymous
    • one year ago
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    u said area under the curve is NOT top minus bottom but thats what i just did ?

  177. anonymous
    • one year ago
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    slow

  178. anonymous
    • one year ago
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    it is always \[F(b)-F(a)\]

  179. anonymous
    • one year ago
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    but if it is the area between two curves you have to subtract the lower one from the upper one BEFORE you integrate

  180. anonymous
    • one year ago
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    so in your example it was only one curve \(y=x^2+1\) from \(0\) to \(4\) you do \[\int_0^2 (x^2+1)dx\]

  181. anonymous
    • one year ago
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    ohh u mean we have to simplify then anti but in area under their is no simplfying to do just straight

  182. anonymous
    • one year ago
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    but if it was between to curves, you have to subtract first, then integrate

  183. anonymous
    • one year ago
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    yes right, what you said

  184. anonymous
    • one year ago
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    now i have a question

  185. anonymous
    • one year ago
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    do you really love homework?

  186. anonymous
    • one year ago
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    according to os you posted this question two hours ago! so i guess you must

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