MHB Cat's question at Yahoo Answers regarding approximate integration

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To approximate the total water consumption from the storage tank over 24 hours, the integral T = ∫(0 to 24) C(t) dt is used, where C(t) = 25e^(-0.05(t-15)^2). Since the integrand lacks an elementary anti-derivative, numerical methods like the Midpoint Rule are applied for approximation. Calculations using Wolfram|Alpha with varying sample sizes yield results around 197.728 thousand gallons. The closest answer from the options provided is c) 197.727.
MarkFL
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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.
 
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Hello Cat,

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

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

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:

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

Now, we find that:

$$\Delta t=\frac{24}{n}$$

$$t_k=\frac{24k}{n}$$

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

Hence:

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

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):

[TABLE="class: grid, width: 200"]
[TR]
[TD]$n$[/TD]
[TD]$T_n$[/TD]
[/TR]
[TR]
[TD]10[/TD]
[TD]197.814[/TD]
[/TR]
[TR]
[TD]100[/TD]
[TD]197.729[/TD]
[/TR]
[TR]
[TD]1000[/TD]
[TD]197.728[/TD]
[/TR]
[TR]
[TD]10000[/TD]
[TD]197.728[/TD]
[/TR]
[/TABLE]

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