Question

calc help !! fan, testimonial , and medal!!!!

Answer 1

@iGreen

Answer 2

@phi i need serious help from like 2 days ago please!!

Answer 3

Since \(R(t)\approx W(t)\) gives the rate of the water flow, the average rate of flow is given by the average value of \(W(t)\). This amounts to settig up and evaluating the following integral:
\[\frac{1}{8-0}\int_0^8W(t)\,dt\]

Answer 4

and i just plug in?

Answer 5

@Compassionate do you know??

Answer 6

@srossd ?

Answer 7

HI!!

Answer 8

assuming @sithsandgiggles is right, here is your answer
http://www.wolframalpha.com/input/?i=%28int+0+to+8++ln%28x^2%2B7%29%29%2F8

Answer 9

\[\int \log(t^2+7)dt\] is not something i would want to compute without a calculator

Answer 10

Yes, you plug in W(t). So it becomes
\[\frac{1}{8}\int\limits_0^8 \ln(t^2+7)\,dt\]
Now, you need the antiderivative. This isn't fun. Here goes:
\[\int \ln(t^2+7)\,dt = t\ln(t^2+7)-\int \frac{2t^2}{t^2+7}\,dt \\ = t\ln(t^2+7)-2\int 1- \frac{7}{t^2+7}\,dt = t\ln(t^2+7)-2t + 2\sqrt{7}\tan^{-1}(t/\sqrt{7})\]
Then you can evalulate that at 0 and 8 and subtract.

Answer 11

That first step was integration by parts, by the way.

Answer 12

3.0904 @srossd ?

Answer 13

what do you mean evaluate at 0 and 8?? @srossd

Answer 14

Yes, 3.0904 is correct. You take the antiderivative, plug in t = 0, subtract that from the value at t =8, and then divide by 8 to get the average.

Answer 15

thank you(:!!