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

There isn't one

Attempting to find the average value using an integral:

is impossible because the integral diverges.

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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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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.

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.

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.

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