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Find the average value of the following function, assuming that the two points a

ID: 3120259 • Letter: F

Question

Find the average value of the following function, assuming that the two points are connected by a line segment. Graphing might help! x 10 15 y 11 17 Compare to the simple arithmetic average of the y values. Find the average value of the following function, assuming that the successive points are connected by line segments. Graphing might help! x 10 15 20 y 11 17 11 Compare (iii) to the simple arithmetic average of the y values in that part. What can you say about how the arithmetic average of the y values relates to the average in a function sort of way? (assuming the horizontal spacing of the points is all equal, as it is in the above examples)

Explanation / Answer

For problem (i)
The avg value of this function will be given by taking the area of the graph and diving by the width; The area = 1/2 * length*height = 1/2*(17-11)(15-10) = 15; width = 15-10 = 5 Thus avg value = 15/5 = 3; This is in addition to y=11 thus the true avg = 11+3 = 14;
Simple arithmetic avg of y values = 11+17 / 2 = 28/2 = 14; Thus in problem (i) the avg value is the same as the simple arithmetic avg of the y values;

For problem (ii)
simple arithmetic avg of the y values = 11+17+11 / 3 = 39/3 = 13;
Avg value of the function: Avg of the graph from x=10 to x=15 is 14 as calcualted in problem (i)
Similarly the avg value of the graph from x=15 to x=20 will also be 14 since the same line is now decreasing with width = 5 and height = 6; Thus the avg of these two averages of 14& 14 will be 14 again.
Thus for problem (ii) we can see that simple arithmetic avg will not be equal to the avg value of the function.

Conclusion: Thus we can conclude that the avg value of the function calculated from area of the function and divided by the width is a true representation of the avg whereas teh arithmetic avg of y values is dependent upon the exact y-values of the points that are chosen. These both methods need not reveal the same value as average.

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