# Heater!

I'm slightly obsessed with mathematics, my cats, and the Dave Matthews Band.

### Business Calculus problem, work included. Why am I coming up with the wrong answer?

The problem is as follows:

"Plutonium-239 has a decay rate of approximately 0.003% per year. Suppose that plutonium-239 is released into the atmosphere each year for 20 years at a constant rate of 1 lb per year. How much plutonium-239 will remain in the atmosphere after 20 years?"

My answer:

The amount of plutonium remaining is given exactly by the definite integral shown, where the chemical is released into the atmosphere at a rate of R(t) pounds per year for T years at a decay rate of k.

∫_(t to 0)(R(t) e^kt (dt)

Substitute the given values in the definition for the future amount. The value of k will be negative because it is decaying.

∫(20 to 0) 1 e^(-0.003t)(dt)

Now, we integrate with respect to t.

[-333.33e^(-0.003(t)) ](20 to 0)

Lastly, we evaluate the result over the integral from t = 0 from t = 20.

-333.33(e^(-0.003(20))-e^(-0.003(0))) ≈ 19.41162802

Therefore, the approximate amount of plutonium-239 in the atmosphere after 20 years is 19.412 lbs.

I'm coming up with the answer of 19.412 lbs, but the book says the answer is 19.994 lbs.

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