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

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

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