If a factory continuously pumps pollutants into the air at the rate of √(t)/15 tons per day, then the amount dumped after 3 days is

a)3.464
b)0.019
c)0.115
d)0.231

Note: I have integral from 0 to 3 of √(t)/15 dt but I guess something is wrong. Can you please, show me your work. Thanks

3 answers

Did you get this?
∫ √(t)/15 dt
= [(2/45 t^(3/2) ] from 0 to 3

I get one of the choices as the answer
Stop disliking the comment. Reiny has it correct. Just solve the integral with the range being 0 to 3 and you'll get the right answer.
the answer is .231
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