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Given that a population (eg. of bacteria) grows exponentially with the function

$$p 2^{\frac{t}{10}}$$

where $p$ is the initial population and $t$ is the time in minutes. Given that the population consumes (eg. food) at a steady rate, how would you calculate how much the population consumes?

Assume that each bacterium consumes $1µg$ of food per minute, how much food would an initial population of 100 bacteria have consumed after 10 minutes? The solution lies somewhere between the consumption of 100 and 200 bacteria over 10 minutes, thus between $1000µg$ and $2000µg$.

I figured I needed an integral to solve this problem. If I'm not mistaken, the integral of the function above is:

$$\int_0^{10} p 2^{\frac{t}{10}} dt = \frac{10 p 2^{\frac{t}{10}}}{\ln{2}}$$

For $p=100$ the result is $2885.39µg$, which is higher than the expected maximum of $2000µg$. Where did I go wrong?

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    +1 for showing what you had done, allowing svenkatr to give a useful response2010-10-21

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You have a definite integral

$\int_{0}^{10} p 2^{\frac{t}{10}} dt = \left[ \frac{10p2^{\frac{t}{10}}}{ln 2} \right]_{0}^{10} = \frac{10p}{ln2} = 1442.69 \mu g $ of food consumed.

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    It looks like you just forgot to subtract the value at the lower limit.2010-10-21
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    I've subtracted the value at the lower limit in my answer. Stijn Van Bael had omitted that in his calculations.2010-10-21
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    yes, I forgot to put in the @Stijn Van Bael. Sorry2010-10-21