Calculating the Average of 2 Functions

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To calculate the average of two functions over a specified period, the correct approach involves integrating each function over its respective interval and then normalizing by the total duration. The first function, defined from 0 to 5 ms, is 1 - e^(-t/1 ms), while the second function from 5 ms to 10 ms is e^(-t/1 ms). The average is computed as \bar{f} = (1/10) * [∫ from 0 to 5 ms of (1 - e^(-t/1 ms)) dt + ∫ from 5 ms to 10 ms of e^(-t/1 ms) dt]. This ensures that both functions are appropriately weighted according to their active periods. The final formula accurately reflects the contributions of both functions over the entire 10 ms interval.
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Homework Statement


My question pertains to taking the average over a particular period that is composed of 2 functions. For example, from [0, 5ms] the function is defined by 1-e^(-t/1ms)and then by e^(-t/1ms) from [5ms, 10ms].

Will the average from 0-->10 simply be the following:
\bar{f}=\frac{1}{5}\int^{5 ms}_{0}{1-e^{-t/1 ms}}+\frac{1}{5}\int ^{5}_{0}{e^{-t/1 ms}}
 
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If you look at the bottom graph that is orange, it gives visual representation as to what I am trying to take the average of. One period, two functions. One function for half the period, another function for the last half.
1873_8RC%20Input%20Waveform.JPG
 
Nvm, pretty sure it is as follows:
\bar{f}=\frac{1}{10}\int^{5 ms}_{0}{1-e^{-t/1 ms}}+\frac{1}{10}\int ^{5}_{0}{e^{-t/1 ms}}
Since the entire period is 10, then both contributions from each function needs to be divided by this number. If I hadn't done this, I would not be considering the other half where each function is zero.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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