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anonymous

  • one year ago

Last Question of the day The function f is continuous on the interval [4, 15], with some of its values given in the table above. Estimate the average value of the function with a Trapezoidal Approximation, using the 4 intervals between those given points. x 4 9 11 14 15 f(x) –6 –11 –18 –21 –25

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  1. anonymous
    • one year ago
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    okay so i'v done this a couple of times but i keep getting the wrong answer.

  2. SolomonZelman
    • one year ago
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    you have a bunch of trapezoids, that is all. Area of each trapezoid is: \(\large\displaystyle A=(\Delta x)(h_1+h_2)/2\) where \(\Delta x\) is the width and \(\large\displaystyle (h_1+h_2)/2\) is the average height \(\large\displaystyle Trapezoid_1=(9-4)(-6+-11)/2\) \(\large\displaystyle Trapezoid_2=(11-9)(-18+-11)/2\) \(\large\displaystyle Trapezoid_3=(14-11)(-21+-18)/2\) \(\large\displaystyle Trapezoid_4=(15-14)(-25+-21)/2\)

  3. SolomonZelman
    • one year ago
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    then add all trapezoids.

  4. SolomonZelman
    • one year ago
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    Dont wonder the area is negative, that is because the entire thing is under the x-axis.

  5. anonymous
    • one year ago
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    yep im getting the same answer i was getting before , still wrong

  6. anonymous
    • one year ago
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    these are my answer choices –12.727 –11.546 –16.273 –13.909

  7. anonymous
    • one year ago
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    @SolomonZelman you still there ?

  8. SolomonZelman
    • one year ago
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    yes, i am here. I am just glitching outside the US a little.

  9. SolomonZelman
    • one year ago
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    what did you get when you added the trapezoids?

  10. anonymous
    • one year ago
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    153, i did the same thing but by taking the left point and right point sum and averaging them

  11. SolomonZelman
    • one year ago
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    Yes, the area of the trapezoid is given by the product of its average height times the width, and the area of your full shape is the sum of all trapezoids... i dont understand why these options are here.

  12. SolomonZelman
    • one year ago
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    i get -153 doing a completely valid method.... maybe your table has typos?

  13. anonymous
    • one year ago
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    perhaps we're finding the wrong thing . it says find the average value might mean the average slope not average area under the curve

  14. anonymous
    • one year ago
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    @ganeshie8 any idea?

  15. freckles
    • one year ago
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    \[\frac{1}{b-a} \int\limits _a^b f(x) dx\]

  16. JoannaBlackwelder
    • one year ago
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    The average function value formula is:|dw:1439414018001:dw|

  17. freckles
    • one year ago
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    is average value of a function on interval [a,b]

  18. SolomonZelman
    • one year ago
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    oh the area, sorry....

  19. SolomonZelman
    • one year ago
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    the average is what they need...

  20. freckles
    • one year ago
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    so you made it this far: \[\frac{1}{15-4} \int\limits_4^{16} f(x) dx \\ \frac{1}{15-4}(-153)\]

  21. freckles
    • one year ago
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    yes that 16 is suppose to be a 15

  22. anonymous
    • one year ago
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    okay im getting 13.909 looks right thank you @freckles

  23. freckles
    • one year ago
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    well -13.909 but yah

  24. anonymous
    • one year ago
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    yep , my bad .

  25. freckles
    • one year ago
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    you should give @solomonzelman a medal he did the harder part of the problem

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