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

Answer 1

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.

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Answer 2

Christopher Allers

To find the average value of the function ( f(x) = 2x \sec^2(x) ) on the interval ([0, \frac{\pi}{4}]), you need to calculate the definite integral of the function over the interval and then divide by the length of the interval.

[ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) , dx ]

In this case, (a = 0) and (b = \frac{\pi}{4}). Substituting these values into the formula:

[ \text{Average value} = \frac{1}{\frac{\pi}{4} - 0} \int_{0}^{\frac{\pi}{4}} 2x \sec^2(x) , dx ]

Now, integrate (2x \sec^2(x)) with respect to (x) over the interval ([0, \frac{\pi}{4}]). After integration, divide by (\frac{\pi}{4}) to find the average value.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.