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Solve the initial value problem $\frac{dy}{dx}=3{x}^{\mathrm{-2}},y\left(1\right)=2.$
$y=-\frac{3}{x}+5$
Initial-value problems arise in many applications. Next we consider a problem in which a driver applies the brakes in a car. We are interested in how long it takes for the car to stop. Recall that the velocity function $v\left(t\right)$ is the derivative of a position function $s\left(t\right),$ and the acceleration $a\left(t\right)$ is the derivative of the velocity function. In earlier examples in the text, we could calculate the velocity from the position and then compute the acceleration from the velocity. In the next example we work the other way around. Given an acceleration function, we calculate the velocity function. We then use the velocity function to determine the position function.
A car is traveling at the rate of $88$ ft/sec $(60$ mph) when the brakes are applied. The car begins decelerating at a constant rate of $15$ ft/sec ^{2} .
Suppose the car is traveling at the rate of $44$ ft/sec. How long does it take for the car to stop? How far will the car travel?
$2.93\phantom{\rule{0.2em}{0ex}}\text{sec},\phantom{\rule{0.2em}{0ex}}64.5\phantom{\rule{0.2em}{0ex}}\text{ft}$
For the following exercises, show that $F\left(x\right)$ are antiderivatives of $f\left(x\right).$
$F\left(x\right)=5{x}^{3}+2{x}^{2}+3x+1,f\left(x\right)=15{x}^{2}+4x+3$
${F}^{\prime}\left(x\right)=15{x}^{2}+4x+3$
$F\left(x\right)={x}^{2}+4x+1,f\left(x\right)=2x+4$
$F\left(x\right)={x}^{2}{e}^{x},f\left(x\right)={e}^{x}\left({x}^{2}+2x\right)$
${F}^{\prime}\left(x\right)=2x{e}^{x}+{x}^{2}{e}^{x}$
$F\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x,f\left(x\right)=\text{\u2212}\text{sin}\phantom{\rule{0.1em}{0ex}}x$
$F\left(x\right)={e}^{x},f\left(x\right)={e}^{x}$
${F}^{\prime}\left(x\right)={e}^{x}$
For the following exercises, find the antiderivative of the function.
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