some typos
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@@ -589,7 +589,7 @@ The answer will either be at a critical point, at $0$ or as $x$ goes to $\infty$
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$$
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[\text{erf}(x)]' = \frac{2}{\pi}e^{-x^2}.
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[\text{erf}(x)]' = \frac{2}{\sqrt{\pi}}e^{-x^2}.
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$$
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Oh, this is never $0$, so there are no critical points. The maximum occurs at $0$ or as $x$ goes to $\infty$. Clearly at $0$, we have $\text{erf}(0)=0$, so the answer will be as $x$ goes to $\infty$.
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@@ -755,7 +755,7 @@ A junior engineer at `Treadmillz.com` is tasked with updating the display of cal
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**********
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```
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In this example display there was 1 calorie burned in the first minute, then 2, then 5, 5, 4, 3, 2, 2, 1. The total is $24$.
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In this example display there was 1 calorie burned in the first minute, then 2, then 5, 5, 4, 4, 3, 2, 2, 1. The total is $29$.
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In her work the junior engineer found this old function for updating the display
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@@ -814,7 +814,7 @@ end
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Then the "area" represented by the dots stays fixed over this time frame.
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The engineer then thought a bit more, as the form of her answer seemed familiar. She decides to parameterize it in terms of $t$ and found with $h=1/n$: `c(t) = (C(t) - C(t-h))/h`. Ahh - the derivative approximation. But then what is the "area"? It is no longer just the sum of the dots, but in terms of the functions she finds that each column represents $c(t)\cdot h$, and the sum is just $c(t_1)h + c(t_2)h + \cdots c(t_n)h$ which looks like an approximate integral.
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The engineer then thought a bit more, as the form of her answer seemed familiar. She decides to parameterize it in terms of $t$ and found with $h=1/n$: `c(t) = (C(t) - C(t-h))/h`. Ahh - the derivative approximation. But then what is the "area"? It is no longer just the sum of the dots, but in terms of the functions she finds that each column represents $c(t)\cdot h$, and the sum is just $c(t_1)h + c(t_2)h + \cdots + c(t_n)h$ which looks like an approximate integral.
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If the display were to reach the modern age and replace LED "dots" with a higher-pixel display, then the function to display would be $c(t) = C'(t)$ and the area displayed would be $\int_{t-10}^t c(u) du$.
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