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anonymous

  • 5 years ago

The rate at which water is sprayed on a field of vegetables is given by R(t)=2sqrt(1+5t^3), where t is in minutes and R(t) is in gallons per minute. During the time interval 0 to 4 inclusive, what is the average rate of water flow, in gallons per minute?

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  1. anonymous
    • 5 years ago
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    The formula is 1/b-a int(equation) from zero to four. Just integrate normally and multiply by the first part.

  2. Regan
    • 5 years ago
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    integrate with limits 0,4 then divide by 4

  3. anonymous
    • 5 years ago
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    Ie, \[1\div4 \int\limits_{0}^{4} 2\sqrt{1+5t ^{3}} dt\]

  4. Regan
    • 5 years ago
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    nailed it

  5. anonymous
    • 5 years ago
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    I got 14.691 What do a & b represent in the average rate equation?

  6. anonymous
    • 5 years ago
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    It is the average value theorem.

  7. anonymous
    • 5 years ago
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    Essentially, to explain geometrically, it is the same as taking the average of any numbers. You take the sum and then divide by how many numbers you had originally, right? You had a certain amount between b, which is 4, and a, which is 2. That is represented by the integration. You divide by their difference because it is the space between in terms of x, or in this case, t. Does that answer your question?

  8. anonymous
    • 5 years ago
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    Why does a=2

  9. anonymous
    • 5 years ago
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    Therefore, the average value theorem states that the average value is simply the integration times 1/b-a

  10. anonymous
    • 5 years ago
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    I'm sorry, that was a typo. A is 0.

  11. anonymous
    • 5 years ago
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    Ok, thanks

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