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At some point in their lives most people will suffer from at least one onset of low back pain. This disorder can trigger excruciating pain and temporary disability, but its causes are hard to diagnose. It is well known that low back pain alters motor trunk patterns; thus it is of interest to study the causes for these alterations and their extent. Due to the different possible causes of this type of pain, a "control" group of people is hard to obtain for laboratory studies. However, pain can be stimulated in healthy people and muscle movement ranges can be compared. Controlled back pain can be induced by injecting saline solution directly into related muscles or ligaments. The transfer function from infusion rate to pain response was obtained experimentally by injecting a 5% saline solution at six different infusion rates over a period of 12 minutes. Subjects verbally rated their pain every 15 seconds on a scale from 0 to 10, with 0 indicating no pain and 10 unbearable pain. Several trials were averaged and the data was fitted to the following transfer function:

G(s)=9.72×108(s+0.0001)(s+0.009)2(s2+0.018s+0.0001)G(s)=\frac{9.72 \times 10^{-8}(s+0.0001)}{(s+0.009)^{2}\left(s^{2}+0.018 s+0.0001\right)}

For experimentation, it is desired to build an automatic dispensing system to make the pain level constant as shown in figure below. It follows that ideally the injection system transfer function has to be


to obtain an overall transfer function M(s)G(s) \approx 1. However, for implementation purposes M(s) must have at least one more pole than zeros. Find a suitable transfer function, M(s) by inverting G(s) and adding poles that are far from the imaginary axis.

Infusion PumpHuman responseConstantConstantInfusionM(s)G(s)backRatepain\scriptscriptstyle \begin{matrix} && \text{Infusion Pump} && \text{Human response}\\ \text{Constant} &&&&&& \text{Constant}\\ \text{Infusion} & \mathrm{\rightarrow} & \text{M(s)} & \mathrm{\rightarrow} & \text{G(s)} & \mathrm{\rightarrow} & \text{back}\\ \text{Rate} &&&&&& \text{pain}\\ \end{matrix}


Find all the antiderivatives of the following function. Check your work by taking derivatives. f(x) = 2 sin x + 1


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A function FF is and antiderivative of ff on an interval II provided F(x)=f(x)F'(x)=f(x), for all xx in II.

In our case, we have f(x)=2sin(x)+1f(x)=2\sin(x)+1. We have to find F(x)F(x), such that F(x)=f(x)F'(x)=f(x). Therefore,

F(x)=2cos(x)+x+C\begin{gather*} F(x)=-2\cos(x)+x+C \end{gather*}

To verify, we calculate F(x)F'(x).

F(x)=(2cos(x)+x+C)F(x)=2sin(x)+1F(x)=f(x)\begin{gather*} F'(x)=(-2\cos(x)+x+C)'\\ F'(x)=2\sin(x)+1\\ F'(x)=f(x) \end{gather*}

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