Cat's question at Yahoo Answers regarding approximate integration

[MarkFL]

Here is the question:

Calculus question: which would best approximate total water consumption of storage tank?


Water is pumped from a storage tank and the flower of water from the tank is given by C(t) = 25e ^-0.05(t-15)^2 thousand gallons per hour, where t is the number of hours since midnight. Which of the following best approximates the total water consumption for one day (in thousands of gallons)?

a) 33.964
b) 164.202
c) 197.727
d) 198.166
e) 202.144

How do you find this answer? Please explain how you work through this problem, thank you!

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Cat,

To find the total consumption $T$ of water for 24 hours, we may state:

\(\displaystyle T=\int_{0}^{24} C(t)\,dt=25\int_{0}^{24} e^{-\frac{(t-15)^2}{20}}\,dt\)

Now, the integrand in this problem does not have an anti-derivative expressible in elementary terms, so we must use either the error function or approximate integration. For simplicity of computation and aided by a computer, let's use the Midpoint Rule and state:

\(\displaystyle T\approx\Delta t\sum_{k=0}^{n-1}\left(C\left(\frac{t_{k}+t_{k+1}}{2} \right) \right)\)

Now, we find that:

\(\displaystyle \Delta t=\frac{24}{n}\)

\(\displaystyle t_k=\frac{24k}{n}\)

\(\displaystyle \frac{t_{k}+t_{k+1}}{2}=\frac{12}{n}(2k+1)\)

Hence:

\(\displaystyle T\approx T_n=\frac{600}{n} \sum_{k=0}^{n-1}\left(\exp\left(-\frac{\left(\dfrac{12}{n}(2k+1)-15 \right)^2}{20} \right) \right)\)

Now, at the site Wolfram|Alpha: Computational Knowledge Engine I used the command:

sum of (600/n)exp(-((12/n)(2k+1)-15)^2/20) for k=0 to n-1

where I substituted powers of 10 for $n$ and obtained (to 3 decimal places):

$n$	$T_n$
10	197.814
100	197.729
1000	197.728
10000	197.728

Thus, the choice given by c) is the closest.