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# 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.

Cidyah, what are you doing to come to those answers?

### 2 Answers

- cidyahLv 71 decade ago
1----0.99997

2----1.9999100009

3----2.999820003599973

4----3.999700008999865

5----4.999550017999595

6----5.999370031499055

7----6.9991600503981095

8----7.9989200755965975

9----8.99865010799433

10----9.99835014849109

11----10.998020197986635

12----11.997660257380696

13----12.997270327572975

14----13.996850409463148

15----14.996400503950865

16----15.995920611935746

17----16.995410734317385

18----17.994870871995357

19----18.994301025869195

20----19.99370119683842

I am not sure whether this is a half-life problem. I computed the values exactly as the problem wanted me to.