Chapter 4: Problem 47
Find the position function \(s(t)\) from the given velocity or acceleration function and initial value(s). Assume that units are feet and seconds. $$v(t)=40-\sin t, s(0)=2$$
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Chapter 4: Problem 47
Find the position function \(s(t)\) from the given velocity or acceleration function and initial value(s). Assume that units are feet and seconds. $$v(t)=40-\sin t, s(0)=2$$
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Evaluate the integral. $$\int_{0}^{1} \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x$$
Evaluate the definite integral. $$\int_{1}^{4} \frac{x-1}{\sqrt{x}} d x$$
Find the average value of the function on the given interval. \(f(x)=x^{2}+2 x,[0,1]\)
Show that \(\int \frac{-1}{\sqrt{1-x^{2}}} d x=\cos ^{-1} x+c\) and \(\int \frac{-1}{\sqrt{1-x^{2}}} d x=-\sin ^{-1} x+c\) Explain why this does not imply that \(\cos ^{-1} x=-\sin ^{-1} x .\) Find an equation relating \(\cos ^{-1} x\) and \(\sin ^{-1} x\)
In this exercise, we guide you through a different proof of \(\lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=1 .\) Start with \(f(x)=\ln x\) and the fact that \(f^{\prime}(1)=1 .\) Using the alternative definition of derivative, we write this as \(f^{\prime}(1)=\lim _{x \rightarrow 1} \frac{\ln x-\ln 1}{x-1}=1 .\) Explain why this implies that \(\lim _{x \rightarrow 1} \frac{x-1}{\ln x}=1 .\) Finally, substitute \(x=e^{h}.\)
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