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## Main Question or Discussion Point

This question popped up in my high school class today. I couldn't think of an explanation that worked for the students, other than "the definition of the average value states that the function has to be continuous on the interval."

Their intuition helped them understand the definition of the average value of f(x): To find the average height of an area, take the area and divide by the width of the area. Thus, the average value of a function f(x) is the integral of f(x) from a to b, divided by b-a, or [tex]\frac{1}{b-a}\displaystyle\int^b_a f(x)\,dx[/tex]

No problem with this... yet. Then, came improper integrals.

They calculated the following integral:

[tex]\displaystyle\int^4_1 \frac{dx}{(x-2)^\frac{2}{3}} = 3+3\sqrt[3]{2}[/tex]

This was followed by the question, "well, even though there's an infinite discontinuity, the area over the interval is finite. Why can't we divide by the width of the interval (3) and get the average value of f(x) over that interval?"

Anyone have a better answer than I had, that they (relatively bright students) might be able to comprehend?

Their intuition helped them understand the definition of the average value of f(x): To find the average height of an area, take the area and divide by the width of the area. Thus, the average value of a function f(x) is the integral of f(x) from a to b, divided by b-a, or [tex]\frac{1}{b-a}\displaystyle\int^b_a f(x)\,dx[/tex]

No problem with this... yet. Then, came improper integrals.

They calculated the following integral:

[tex]\displaystyle\int^4_1 \frac{dx}{(x-2)^\frac{2}{3}} = 3+3\sqrt[3]{2}[/tex]

This was followed by the question, "well, even though there's an infinite discontinuity, the area over the interval is finite. Why can't we divide by the width of the interval (3) and get the average value of f(x) over that interval?"

Anyone have a better answer than I had, that they (relatively bright students) might be able to comprehend?