+1904 
Anyone who know how to answer this question and can give step by step instructions, I would really appreciate it. Thanks.
+37153 \(\int_{0}^{40}\)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?
