Integral of a function + limit

[Bikkehaug]

Homework Statement

Plutonium is a radioactive waste.. etc. A mass og 1.000 kg will after x years be reduced to:

m(x) = 1.000 * e^-2.89*10-5x

The yearly Waste of plutonium is 1kg. The total plutonium mass after N years is given by:

∫m(x)dx where the upper value of the integral is N, and the lower value is 0 (not sure how to Write this directly into the formula)

Compute the integral, and estimate the total waste of plutonium after a long time.

Homework Equations

The Attempt at a Solution

Solving the integral: ∫1.000 * e^-2.89*10-5x dx

= [$\frac{1}{-2.89*10^-5}$ * e^-2.89*10-5x ] Again, the higher value is N, the lower value is 0.

If I plot a graph of this Equation, I get a linear graph With the mass after, say 5000 years is 5000kg. After 3 years the mass is 3kg.

Does anyone have any idea about how I can proceed from this?

 

[jbunniii]

Try plotting over a longer time scale (millions of years). Unfortunately, $5000$ years is negligible due to the $-2.89\times 10^{-5}$ coefficient. One way to understand why you are seeing a straight line is to write $\alpha = -2.89\times 10^{-5}$. Then for values of $x$ such that $\alpha x$ is small, we can approximate $e^{ax} \approx 1 + ax$, so the result of your integration can be approximated as
$$\left.\frac{1}{\alpha}(1 + ax)\right|_{0}^{N} = \left.\left(\frac{1}{\alpha} + x\right)\right|_{0}^{N} = N$$
So the growth is approximately linear until $x$ is large enough that this approximation no longer holds.

 

[pasmith]

Homework Statement

Plutonium is a radioactive waste.. etc. A mass og 1.000 kg will after x years be reduced to:

m(x) = 1.000 * e^-2.89*10-5x

The yearly Waste of plutonium is 1kg. The total plutonium mass after N years is given by:

∫m(x)dx where the upper value of the integral is N, and the lower value is 0 (not sure how to Write this directly into the formula)

Compute the integral, and estimate the total waste of plutonium after a long time.

Homework Equations

The Attempt at a Solution

Solving the integral: ∫1.000 * e^-2.89*10-5x dx

= [$\frac{1}{-2.89*10^-5}$ * e^-2.89*10-5x ] Again, the higher value is N, the lower value is 0.

So that's
$$\frac{1}{2.89 \times 10^{-5}} \left(1 - e^{-2.89 \times 10^{-5} N}\right).$$

Does anyone have any idea about how I can proceed from this?

"estimate ... after a long time" means "let $N \to \infty$".