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jocelynevsq

idenify the definite integral that represents the area of the region bounded by the graphs of y=x and y=5x-x^3

  • one year ago
  • one year ago

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  1. cherio12
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    try graphing them first to determine which function is the top and which is the bottom

    • one year ago
  2. jocelynevsq
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    ok

    • one year ago
  3. jocelynevsq
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    still don't get it

    • one year ago
  4. jocelynevsq
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    |dw:1359920225466:dw|

    • one year ago
  5. zepdrix
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    So ummmm yah that graph looks accurate. I always get confused by these types of problems. When the upper and lower functions switch places. But I think we can figure it out. Hmm.

    • one year ago
  6. zepdrix
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    Let's start by finding points of intersection.\[\large y=x \qquad y=5x-x^3\qquad \qquad \rightarrow \qquad \qquad x=5x-x^3\]

    • one year ago
  7. zepdrix
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    Which turn out to be, \(\large x=0, \quad x=-2, \quad x=2\).

    • one year ago
  8. zepdrix
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    Let's look at just the right side for a moment, maybe it will make a little more sense that way.

    • one year ago
  9. zepdrix
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    |dw:1359921446750:dw|See how I've drawn a small rectangle? Let's find the area of that rectangle.

    • one year ago
  10. zepdrix
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    The area will be `base x height`. The base has a thickness of \(dx\). The height is going to be `the upper function` minus `the lower function`.

    • one year ago
  11. zepdrix
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    \[\large A=\left(height\right)(base)\]\[\large A=\left(5x-x^3-x\right)(dx)\]See how the line y=x is BELOW the other one? So we subtract y=x from the other function to get the correct height. So this represents the area of a small rectangular slice. Now we want to Integrate (sum up) all of the rectangles between these two curves.

    • one year ago
  12. zepdrix
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    And we'll include our intersecting points.

    • one year ago
  13. zepdrix
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    \[\large \int\limits_0^2 4x-x^3 \; dx\] This represents the area between the two curves on the RIGHT side. We still need to find the area of the other chunk. Notice how they switch places on the left? The line y=x is now the UPPER function! So we would subtract y=5x-x^3 from the line y=x. And our limits of integration would be a little different.

    • one year ago
  14. zepdrix
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    There is a little trick we could pull though c: See how they're both odd functions? They're both symmetric about the origin. So the area on the left will be the same as the area on the right. We could simply multiply this integral by 2.\[\huge A_{total}\qquad =\qquad 2\int\limits\limits_0^2 4x-x^3 \; dx\]

    • one year ago
  15. jocelynevsq
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    oh okay i see how you got that. thanks!

    • one year ago
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