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Anyone who know how to answer this question and can give step by step instructions, I would really appreciate it.  Thanks.

 Feb 4, 2019
 #1
avatar+19811 
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\(\int_{0}^{40}\)c(t) dt      is the total calories burned during the workout......

 

\(\int_{0}^{10}\) t-.05t^2+15  dt     +     \(\int_{10}^{30}\) 20   dt   +       \(\int_{30}^{40}\)3t-.05t^2-25  dt     = total calories burned

 

First part:   1/2 t^2 - .05/3 t^3 + 15 t       From 0 to 10 =    50 - 16.67 + 150  =183.33 cal

 

Second part:   20 t      from 10 to 30 =      600 - 200 = 400 cal

 

Third part:   Think you can do this one?

 Feb 4, 2019

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