anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
okay so i'v done this a couple of times but i keep getting the wrong answer.
SolomonZelman
  • SolomonZelman
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 \(\large\displaystyle Trapezoid_1=(9-4)(-6+-11)/2\) \(\large\displaystyle Trapezoid_2=(11-9)(-18+-11)/2\) \(\large\displaystyle Trapezoid_3=(14-11)(-21+-18)/2\) \(\large\displaystyle Trapezoid_4=(15-14)(-25+-21)/2\)
SolomonZelman
  • SolomonZelman
then add all trapezoids.

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SolomonZelman
  • SolomonZelman
Dont wonder the area is negative, that is because the entire thing is under the x-axis.
anonymous
  • anonymous
yep im getting the same answer i was getting before , still wrong
anonymous
  • anonymous
these are my answer choices –12.727 –11.546 –16.273 –13.909
anonymous
  • anonymous
@SolomonZelman you still there ?
SolomonZelman
  • SolomonZelman
yes, i am here. I am just glitching outside the US a little.
SolomonZelman
  • SolomonZelman
what did you get when you added the trapezoids?
anonymous
  • anonymous
153, i did the same thing but by taking the left point and right point sum and averaging them
SolomonZelman
  • SolomonZelman
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.
SolomonZelman
  • SolomonZelman
i get -153 doing a completely valid method.... maybe your table has typos?
anonymous
  • anonymous
perhaps we're finding the wrong thing . it says find the average value might mean the average slope not average area under the curve
anonymous
  • anonymous
@ganeshie8 any idea?
freckles
  • freckles
\[\frac{1}{b-a} \int\limits _a^b f(x) dx\]
JoannaBlackwelder
  • JoannaBlackwelder
The average function value formula is:|dw:1439414018001:dw|
freckles
  • freckles
is average value of a function on interval [a,b]
SolomonZelman
  • SolomonZelman
oh the area, sorry....
SolomonZelman
  • SolomonZelman
the average is what they need...
freckles
  • freckles
so you made it this far: \[\frac{1}{15-4} \int\limits_4^{16} f(x) dx \\ \frac{1}{15-4}(-153)\]
freckles
  • freckles
yes that 16 is suppose to be a 15
anonymous
  • anonymous
okay im getting 13.909 looks right thank you @freckles
freckles
  • freckles
well -13.909 but yah
anonymous
  • anonymous
yep , my bad .
freckles
  • freckles
you should give @solomonzelman a medal he did the harder part of the problem

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