Question

If a factory continuously pumps pollutants into the air at the rate of the quotient of the square root of t and 15 tons per day, then the amount dumped after 3 days isAll that is needed is to calculate an integral, I'd be happy to help as much as I can.

Answer 1

Step 1

To solve the problem, we need to integrate the given expression and evaluate using the given time.

`\int_0^3(√(t))/(15)dt`
`\int_0^3(1)/(15)(t)^{(1)/(2)}dt`
`=(1)/(15)(\int_0^3(t)^{(1)/(2)})dt`
`\begin{gathered} Apply\text{ the power rule} \\ =(1)/(15)\left[(2)/(3)t^{(3)/(2)}\right]^3_0 \\ Compute\text{ the boundaries} \\ =(1)/(15)\left[(2)/(3)(3)^{(3)/(2)}\right]_^-(1)/(15)\left[(2)/(3)(0)^{(3)/(2)}\right]_^ \\ =(1)/(15)((2)/(3)(3)(\:3^{(1)/(2)}))-0 \end{gathered}`
`\begin{gathered} =(2)/(15)(√(3)) \\ =(2√(3))/(15) \\ \end{gathered}`
`\begin{gathered} \int_0^3(√(t))/(15)dt=(2√(3))/(15)=0.2309401077 \\ \int_0^3(√(t))/(15)dt=0.231 \end{gathered}`

Answer; The amount dumped after 3 days is

`0.231`