- #1

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during the time interval 0<t<4, what is the average rate of water flow??

I'm confused whether to use the formula

f(4)-f(0) / (4-0)

OR

[tex]\frac{1}{4}[/tex][tex]\int[/tex]f(t) evaluated from 0 to 4.

help!!

- Thread starter fiziksfun
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- #1

- 78

- 0

during the time interval 0<t<4, what is the average rate of water flow??

I'm confused whether to use the formula

f(4)-f(0) / (4-0)

OR

[tex]\frac{1}{4}[/tex][tex]\int[/tex]f(t) evaluated from 0 to 4.

help!!

- #2

tiny-tim

Science Advisor

Homework Helper

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Hi fiziksfun !

Hint: Suppose f(t) is a constant, C. Then f(4) = f(0) = C.

So which formula is right?

Hint: Suppose f(t) is a constant, C. Then f(4) = f(0) = C.

So which formula is right?

- #3

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i'm kind of slow, so i still don't understand :[

can you help me more??

can you help me more??

- #4

tiny-tim

Science Advisor

Homework Helper

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If f(t) = C, a constant, then obviously the average of f(t) is C.

But f(4)-f(0) / (4-0) = (C - C)/4 = 0, which obviously is wrong.

And ∫f(t)/4 = ∫C/4 (evaluated from t = 0 to 4) = C, which equally obviously is right!

This works for *any* f(t), because f(t) is the rate of water, so ∫f(t) is the total water.

And so the average rate of water = total/time = ∫f(t)/4.

And so the average rate of water = total/time = ∫f(t)/4.

- #5

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ah, i think i understand, thank you!!!

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