Cesium 137 (Cs¹³⁷)is a Short-Lived Radioactive Isotope $e^{-\left(\frac{\ln 2}{30}\right) t}$ )As a Result of Its Operations, a Nuclear Power Plant

Cesium 137 (Cs¹³⁷) is a short-lived radioactive isotope.It decays at a rate proportional to the amount of itself present and has a half-life of 30 years (i.e., the amount of Cs¹³⁷ remaining t years after A₀ mg of the radioactive isotope is released is given by $e^{-\left(\frac{\ln 2}{30}\right) t}$ ) .As a result of its operations, a nuclear power plant releases Cs¹³⁷ at a rate of 0.12 mg per year.The plant began its operations in 1990, which we will designate as t = 0.Assume there is no other source of this particular isotope.Which of the following integrals give the total amount of Cs¹³⁷ T years after the plant began operations?

A) $\int_{0}^{T} e^{\left(-\frac{0.12 \ln 2}{30}\right) t} d t$
B) $\int_{0}^{T} 0.12 e^{\left(-\frac{\ln 2}{30}\right) t} d t$
C) $\int_{0}^{0.12} \ t e^{\left(-\frac{\ln 2}{30}\right) t} d t$
D) $\left.\int_{0}^{T}\left(e^{\left(-\frac{\ln 2}{30}\right.}\right) t+0.12\right) d t$