Question

Radioactive Waste The rate at which radioactive waste is entering the atmosphere at time t is decreasing and is given by Pe^-kt, where P is the initial rate. Use the improper integral
∫^[infinity]_0 Pe^-kt dt
with P = 50 to find the total amount of the waste that will enter the atmosphere for each value of k.
k=0.04

Answer 1

Answer:

$\displaystyle{\int^{\infty}_0 {50e^{-0.04t}}\, dt}=1250$

Step-by-step explanation:

Given:

$T =\displaystyle{\int^{\infty}_0 {Pe^{-kt}} \, dt}$

where,

T = total amount of waste

P = 50, the initial rate

k = 0.04

t = time

$T =\displaystyle{\int^{\infty}_0 {50e^{-0.04t}} \, dt}$

now we need to solve this integral!

$T =\displaystyle{50\int^{\infty}_0 {e^{-0.04t}} \, dt}$

$T = \left|50\left(\dfrac{e^{-0.04t}}{-0.04}\right)\right|^{\infty}_0$

$T = \left|-1250e^{-0.04t}\right|^{\infty}_0$

$T = (-1250e^{-0.04(\infty)})-(-1250e^{-0.04(0)})$

when any number has a power of negative infinity it is 0. because: $a^-{\infty} = \dfrac{1}{a^{\infty}} = \dfrac{1}{\infty} = 0$, like something being divided by a very large number!

$T = (-1250(0))-(-1250e^0)$

$T = 1250$

this is the total amount of waste