What is the average value of a function #f(x) = 2x sec2 x# on the interval #[0, pi/4]#?

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Answer 1 · 2017-08-19T18:04:20

There isn't one

Recall that #secx# has a vertical asymptote at #x=pi/2#.

Thus, #2xsec(2x)# has a vertical asymptote at #x=pi/4# and there will be no average value for the function on the requested interval.

Attempting to find the average value using an integral:

#1/(pi//4-0)int_0^(pi//4)2xsec(2x)dx#

is impossible because the integral diverges.