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  1. anonymous
    • one year ago
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    Write \[\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(2+k*(5/n))^3 *5/n\] as a definite Integral

  2. anonymous
    • one year ago
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    Find \[dx/dy \int\limits_{2}^{x^3}\ln(x^2)dx\]

  3. anonymous
    • one year ago
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    Find the area bounded by the curves y^2= 2x+6 and x=y+1. Work must be an integral with one variable

  4. anonymous
    • one year ago
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    Write the integral in one variable to find the volume of the solid obtained by rotating the first quadrant region bounded by y=0.5x^2 and y=x about the line x=7.

  5. anonymous
    • one year ago
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    0.5x² = x 0.5x² - x = 0 x (0.5x - 1) = 0 x = 0, 2 Those are the x-coordinates of the two points of intersection of y = 00.5x² and y=x. The full points are (0,0) and (2,2). Revolving the region around a vertical axis like x = 3 is better suited to the cylindrical shell method so that you don't have to solve for y. You integrate the lateral surface area of a cylindrical shell having its height vertically within the bounded region. The radius is 3-x (because x is the distance from the y-axis to the outer edge of the shell, leaving you with 3-x as the distance from the outer edge of the shell to x=3). The height of the cylinder is upper curve minus lower curve = x - 0.5x². Therefore, volume = 2π∫rh dx = 2π∫ (3-x)(x - 0.5x²) dx, from x = 0 to 2 = 2π∫ (3x-x² - 1.5x² + 0.5x³) dx, from x = 0 to 2 = 2π∫ (3x - 2.5x² + 0.5x³) dx, from x = 0 to 2 = 2π [3x²/2 - 2.5x³/3 + 0.5x⁴/4], from x = 0 to 2 = 2π [6 - 20/3 + 2] = 2π [4/3] = 8π/3

  6. anonymous
    • one year ago
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    A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as -32 feet per second squared.

  7. Michele_Laino
    • one year ago
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    question #2: the integral is, integrating by parts, is: \[\Large \begin{gathered} \int {dx\ln \left( {{x^2}} \right)} = x\ln \left( {{x^2}} \right) - \int {x\frac{{dx}}{{{x^2}}}} 2x = \hfill \\ \hfill \\ = x\ln \left( {{x^2}} \right) - 2x \hfill \\ \end{gathered} \]

  8. Michele_Laino
    • one year ago
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    so the definite integral is: \[\Large \left. {x\ln \left( {{x^2}} \right) - 2x} \right|_2^{{x^3}} = {x^3}\ln \left( {{x^6}} \right) - 2{x^3} - 2\ln 4 + 4\]

  9. anonymous
    • one year ago
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    for iluvhomewurk's did they just confuse 3 with 7?

  10. Michele_Laino
    • one year ago
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    Question #3 we can rewrite your equations as follows: \[\Large \left\{ \begin{gathered} x = \frac{{{y^2} - 6}}{2} \hfill \\ x = y + 1 \hfill \\ \end{gathered} \right.\]

  11. Michele_Laino
    • one year ago
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    the intersection point are: \[\begin{gathered} P = \left( { - 1, - 2} \right) \hfill \\ Q = \left( {5,4} \right) \hfill \\ \end{gathered} \]

  12. Michele_Laino
    • one year ago
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    so the requested area, is given by the subsequent integral: \[\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} \]

  13. anonymous
    • one year ago
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    are you still talking about question 3?

  14. Michele_Laino
    • one year ago
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    yes!

  15. Michele_Laino
    • one year ago
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    here is the next step: \[\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {y - \frac{{{y^2}}}{2} + 4} \right|_{ - 2}^4 = 28\]

  16. Michele_Laino
    • one year ago
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    |dw:1433964888030:dw|

  17. anonymous
    • one year ago
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    so the answer is 28?

  18. Michele_Laino
    • one year ago
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    yes!

  19. Michele_Laino
    • one year ago
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    oops.. it is 18

  20. Michele_Laino
    • one year ago
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    sorry!

  21. Michele_Laino
    • one year ago
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    \[\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {y - \frac{{{y^2}}}{2} + 4} \right|_{ - 2}^4 = 18\]

  22. anonymous
    • one year ago
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    its okay Is the work that iluvhomework did correct?

  23. Michele_Laino
    • one year ago
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    please wait I'm checking...

  24. anonymous
    • one year ago
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    okay thank you. I think it is just its 7 instead of 3

  25. anonymous
    • one year ago
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    and i got 0 not 18

  26. Michele_Laino
    • one year ago
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    area=0?

  27. anonymous
    • one year ago
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    the step is to plug in 4 into equation then subtract the equation with -2 pulgged into it correct?

  28. Michele_Laino
    • one year ago
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    yes! correct!

  29. anonymous
    • one year ago
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    When i did that i got 0

  30. Michele_Laino
    • one year ago
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    I'm checking....

  31. Michele_Laino
    • one year ago
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    you are right, I gave you the integrand function, please here is the integral: \[\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {\frac{{{y^2}}}{2} - \frac{{{y^3}}}{6} + 4y} \right|_{ - 2}^4 = 18\]

  32. Michele_Laino
    • one year ago
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    \[\large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {\frac{{{y^2}}}{2} - \frac{{{y^3}}}{6} + 4y} \right|_{ - 2}^4 = 18\]

  33. Michele_Laino
    • one year ago
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    question #3 I make this traslation: \[\Large \left\{ \begin{gathered} x = X + 7 \hfill \\ y = Y \hfill \\ \end{gathered} \right.\] where X, and Y are the new coordinates

  34. Michele_Laino
    • one year ago
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    as you can see my new reference system has the Y-axis along the axis of rotation

  35. Michele_Laino
    • one year ago
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    |dw:1433966377727:dw|

  36. Michele_Laino
    • one year ago
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    so the equations of your lines, becomes: \[\Large \begin{gathered} y = \frac{{{x^2}}}{2} \to Y = \frac{{{{\left( {X + 7} \right)}^2}}}{2} \hfill \\ \hfill \\ y = x \to Y = X + 7 \hfill \\ \end{gathered} \]

  37. anonymous
    • one year ago
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    what problem is this for? the one that ilovehomework started?

  38. Michele_Laino
    • one year ago
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    here is our situation: |dw:1433966625656:dw|

  39. Michele_Laino
    • one year ago
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    yes!

  40. anonymous
    • one year ago
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    Im confused on what work i am supposed to show

  41. anonymous
    • one year ago
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    is ilovehomeworks work not correct?

  42. Michele_Laino
    • one year ago
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    I give my solution, after that you can choose the solution which you prefer

  43. Michele_Laino
    • one year ago
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    I got a different result

  44. anonymous
    • one year ago
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    okay then please outline step by step on how you did the problem. Tnak you

  45. anonymous
    • one year ago
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    okay then please outline step by step on how you did the problem. Tnak you

  46. Michele_Laino
    • one year ago
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    ok! Now our volume is given integrating the area of annulus like below, along the Y-axis: |dw:1433967160201:dw|

  47. anonymous
    • one year ago
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    okay how do you show that mathematically.

  48. Michele_Laino
    • one year ago
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    here is the integral which gives the requested volume: \[\Large \int_0^2 {dY\left\{ {{{\left( { - 7 + Y} \right)}^2} - {{\left( { - 7 + \sqrt {2Y} } \right)}^2}} \right\}} \pi \]

  49. anonymous
    • one year ago
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    so thats my first step to write down?

  50. Michele_Laino
    • one year ago
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    your first step are the equations of traslation, I think

  51. anonymous
    • one year ago
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    okay

  52. Michele_Laino
    • one year ago
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    now, developing that integral, we get: \[\Large \begin{gathered} \int_0^2 {dY\left\{ {{{\left( { - 7 + Y} \right)}^2} - {{\left( { - 7 + \sqrt {2Y} } \right)}^2}} \right\}} \pi = \hfill \\ \hfill \\ = \pi \left( {\left. {\frac{{{Y^3}}}{3} - 16\frac{{{Y^2}}}{2} + 14\sqrt 2 \times \frac{{2{Y^{3/2}}}}{3}} \right|_0^2} \right) \hfill \\ \end{gathered} \]

  53. Michele_Laino
    • one year ago
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    and the final result is: \[\Large V = 8\pi \]

  54. anonymous
    • one year ago
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    A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as -32 feet per second squared.

  55. Michele_Laino
    • one year ago
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    here we have to use the subsequent vector equation: \[\Large {\mathbf{OP}}\left( t \right) = {\mathbf{O}}{{\mathbf{P}}_{\mathbf{0}}} + {{\mathbf{v}}_{\mathbf{0}}}t + \frac{1}{2}{\mathbf{g}}{t^2}\]

  56. Michele_Laino
    • one year ago
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    |dw:1433968218028:dw|

  57. anonymous
    • one year ago
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    okay

  58. Michele_Laino
    • one year ago
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    furthermore, we have to consider this one: \[\Large {\mathbf{v}}\left( t \right) = {{\mathbf{v}}_{\mathbf{0}}} + {\mathbf{g}}t\] |dw:1433968355520:dw|

  59. anonymous
    • one year ago
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    so we use both equations?

  60. Michele_Laino
    • one year ago
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    yes!

  61. Michele_Laino
    • one year ago
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    now I rewrite both equations using z-components: \[\Large \left\{ \begin{gathered} z\left( t \right) = {z_0} + {v_0}t - \frac{1}{2}g{t^2} \hfill \\ {v_z}\left( t \right) = {v_0} - gt \hfill \\ \end{gathered} \right.\]

  62. Michele_Laino
    • one year ago
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    when the ball reaches its maximum height, then its velocity is zero, so we can write: \[\Large 0 = {v_0} - g\tau \] where \tau is the time at which our ball reaches its maximum height

  63. Michele_Laino
    • one year ago
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    so: \[\Large \tau = \frac{{{v_0}}}{g}\]

  64. anonymous
    • one year ago
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    so t=40/-32?

  65. Michele_Laino
    • one year ago
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    no, it is \tau = 40/32=...

  66. Michele_Laino
    • one year ago
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    g is the magnitude of gravity: g=9.81 m/sec^2=32 feet/sec^2

  67. anonymous
    • one year ago
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    wont it be -32?

  68. Michele_Laino
    • one year ago
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    no, since I wrote my scalar equations using the magnitudes or length of the respective vectors the maximum height is then (substituting \tau into the first scalar equation): \[\Large \begin{gathered} z\left( \tau \right) = {z_0} + {v_0}\tau - \frac{1}{2}g{\tau ^2} = h + \frac{{v_0^2}}{g} - \frac{1}{2}\frac{{v_0^2}}{g} = \hfill \\ \hfill \\ = h + \frac{{v_0^2}}{{2g}} \hfill \\ \end{gathered} \]

  69. Michele_Laino
    • one year ago
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    where h=500 feet

  70. Michele_Laino
    • one year ago
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    next, in order to find the requested velocity, we can apply the formula of Torricelli, like below: \[\Large v = \sqrt {2gz\left( \tau \right)} = \sqrt {2gh + v_0^2} \]

  71. Michele_Laino
    • one year ago
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    substituting your data we find: \[\Large \begin{gathered} v = \sqrt {2gz\left( \tau \right)} = \sqrt {2gh + v_0^2} = \sqrt {2 \times 32 \times 500 + {{40}^2}} = \hfill \\ = 183.3\;{\text{feet/sec}} \hfill \\ \end{gathered} \]

  72. Michele_Laino
    • one year ago
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    \[\large \begin{gathered} v = \sqrt {2gz\left( \tau \right)} = \sqrt {2gh + v_0^2} = \sqrt {2 \times 32 \times 500 + {{40}^2}} = \hfill \\ = 183.3\;{\text{feet/sec}} \hfill \\ \end{gathered} \]

  73. anonymous
    • one year ago
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    and thats the final answer?

  74. Michele_Laino
    • one year ago
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    yes!

  75. anonymous
    • one year ago
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    Thank you so much. Could you help me with about 3 more?

  76. Michele_Laino
    • one year ago
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    ok!

  77. anonymous
    • one year ago
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    Use the midpoint rule with n=4 to approximate the area of the region bounded by y=x^3 and y=x

  78. Michele_Laino
    • one year ago
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    please wait I'm pondering...

  79. anonymous
    • one year ago
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    okay thank you

  80. Michele_Laino
    • one year ago
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    the intersection points between those equations are: \[\begin{gathered} P = \left( { - 1, - 1} \right) \hfill \\ Q = \left( {1,1} \right) \hfill \\ O = \left( {0,0} \right) \hfill \\ \end{gathered} \]

  81. Michele_Laino
    • one year ago
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    |dw:1433970078818:dw|

  82. anonymous
    • one year ago
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    ok i see the intersection points

  83. Michele_Laino
    • one year ago
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    so, it is suffice to compute the bounded area which lies into the first quadrant

  84. anonymous
    • one year ago
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    yes I agree with that

  85. Michele_Laino
    • one year ago
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    |dw:1433970288453:dw|

  86. Michele_Laino
    • one year ago
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    the area under the cubic parabola, is given by the subsequent formula: \[\Large A = \frac{{\left( {1 - 0} \right)}}{2}\left( {f\left( {1/4} \right) + f\left( {3/4} \right)} \right)\] where f(x) = x^3

  87. Michele_Laino
    • one year ago
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    so we get: \[\Large \begin{gathered} A = \frac{{\left( {1 - 0} \right)}}{2}\left( {f\left( {1/4} \right) + f\left( {3/4} \right)} \right) = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {\frac{1}{{64}} + \frac{{27}}{{64}}} \right\} = \frac{{14}}{{64}} \hfill \\ \end{gathered} \]

  88. anonymous
    • one year ago
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    simplidy it to 7/32?

  89. Michele_Laino
    • one year ago
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    |dw:1433970540095:dw| the area of bounded region is: \[\Large {A_B} = \frac{1}{2} - \frac{{14}}{{64}} = \frac{7}{{16}}\]

  90. anonymous
    • one year ago
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    oh okay

  91. Michele_Laino
    • one year ago
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    so the requested area is: \[\Large 2{A_B} = \frac{7}{8}\]

  92. anonymous
    • one year ago
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    who do you format it like that?

  93. anonymous
    • one year ago
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    why

  94. Michele_Laino
    • one year ago
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    since our drawing is symmetric with respect to the y-axis: |dw:1433970857126:dw|

  95. anonymous
    • one year ago
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    wouldnt one of the areas be negative because it is below the x axis?

  96. anonymous
    • one year ago
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    so it would be 0?

  97. Michele_Laino
    • one year ago
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    no, since an area is the measure of the subset of the x,y-plane, and the measure of a set, as it is defined by mathematicians, is always positive

  98. anonymous
    • one year ago
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    okay so the fonal answer would be 7/8?

  99. Michele_Laino
    • one year ago
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    yes!

  100. anonymous
    • one year ago
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    okay heres the next one

  101. Michele_Laino
    • one year ago
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    ok!

  102. anonymous
    • one year ago
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    I have completed it i just wanna make sure i did it correct.

  103. anonymous
    • one year ago
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    Trash is delivered to a dumpsite between the hours of midnight(t=0) and 6am (t=6). Selected Values of the rates of deleivery are shown in the table below with t measured in hours and r(t) measured in tons per hours. Use 4 trapezoids to estimate total amount of trash. Round to nearest ton

  104. anonymous
    • one year ago
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    t 0 1 3 5 6 R(t) 4 2 3 2 1

  105. anonymous
    • one year ago
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    Lsum= (1)(4)+(2)(2)+(2)(3)+(1)(2)=4+4+6+2= 16 Rsum= (1)(2)+(2)(3)+(2)(2)+(1)(1)= 2+6+4+1= 13 Tsum= ½(Rsum+Lsum)= ½(16+13)=1/2(29)= 14.5 15 Tons

  106. anonymous
    • one year ago
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    I think i did it correctly

  107. Michele_Laino
    • one year ago
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    please wait, I'm checking your computation...

  108. anonymous
    • one year ago
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    Lsum= (1)(4)+(2)(2)+(2)(3)+(1)(2)=4+4+6+2= 16 Rsum= (1)(2)+(2)(3)+(2)(2)+(1)(1)= 2+6+4+1= 13 Tsum= ½(Rsum+Lsum)= ½(16+13)=1/2(29)= 14.5 15 Tons

  109. anonymous
    • one year ago
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    sorry dont know why it sent again

  110. Michele_Laino
    • one year ago
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    Lsum is right for me whereas I write Rsum like this: Rsum= (2)(2)+(2)(3)+(2)(2)+(1)(1)= 4+6+4+1= 15

  111. anonymous
    • one year ago
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    okat thank you for al this help just two more!

  112. Michele_Laino
    • one year ago
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    ok!

  113. anonymous
    • one year ago
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    Show that the function \[F(x)= \int\limits_{x}^{3x}1/t dt\] is constant on the interval(0, infty)

  114. Michele_Laino
    • one year ago
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    it is simple, since we have: \[\Large \int_x^{3x} {\frac{{dt}}{t}} = \ln \left( {3x} \right) - \ln \left( x \right) = \ln \left( {\frac{{3x}}{x}} \right) = \ln 3\]

  115. anonymous
    • one year ago
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    how did you get ln(3x) and ln(x) from the integral?

  116. Michele_Laino
    • one year ago
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    yes! since we can write: \[\Large \int {\frac{{dt}}{t}} = \ln t + k\]

  117. anonymous
    • one year ago
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    oh okay!

  118. anonymous
    • one year ago
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    and since it comes out to be a number that means it is constant?

  119. Michele_Laino
    • one year ago
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    correct!

  120. anonymous
    • one year ago
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    perfect! last question

  121. Michele_Laino
    • one year ago
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    ok!

  122. anonymous
    • one year ago
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    write\[\lim_{n \rightarrow \infty} \sum_{k=1}^{n} (2+k*5/n)^3*5/n\] as a definite integral

  123. Michele_Laino
    • one year ago
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    I'm pondering...

  124. anonymous
    • one year ago
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    Thank you

  125. Michele_Laino
    • one year ago
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    what is k

  126. anonymous
    • one year ago
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    constant

  127. Michele_Laino
    • one year ago
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    is it a parameter?

  128. Michele_Laino
    • one year ago
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    ok!

  129. anonymous
    • one year ago
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    I think i got it

  130. anonymous
    • one year ago
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    Except 2+x instead of 2+5x

  131. anonymous
    • one year ago
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    does that look correct?

  132. Michele_Laino
    • one year ago
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    here is my answer: I write the sum as integral, like below: \[\Large \int_0^n {dk} {\left( {2 + \frac{{5k}}{n}} \right)^3}\frac{5}{n}\]

  133. anonymous
    • one year ago
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    Thank you for all your help i am finished

  134. Michele_Laino
    • one year ago
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    which is equal to: \[\Large \int_0^n {dk} {\left( {2 + \frac{{5k}}{n}} \right)^3}\frac{5}{n} = \left. {{{\left( {2 + \frac{{5k}}{n}} \right)}^4}\frac{1}{4}} \right|_0^n\]

  135. Michele_Laino
    • one year ago
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    and finally, I get: \[\Large \begin{gathered} \int_0^n {dk} {\left( {2 + \frac{{5k}}{n}} \right)^3}\frac{5}{n} = \left. {{{\left( {2 + \frac{{5k}}{n}} \right)}^4}\frac{1}{4}} \right|_0^n = \hfill \\ \hfill \\ = \frac{{{7^4} - {2^4}}}{4} \hfill \\ \end{gathered} \]

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