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$f\left(x\right)=\frac{1}{{x}^{2}}+x$
$f\left(x\right)={e}^{x}-3{x}^{2}+\text{sin}\phantom{\rule{0.1em}{0ex}}x$
$F\left(x\right)={e}^{x}-{x}^{3}-\text{cos}\left(x\right)+C$
$f\left(x\right)={e}^{x}+3x-{x}^{2}$
$f\left(x\right)=x-1+4\phantom{\rule{0.1em}{0ex}}\text{sin}\left(2x\right)$
$F\left(x\right)=\frac{{x}^{2}}{2}-x-2\phantom{\rule{0.1em}{0ex}}\text{cos}\left(2x\right)+C$
For the following exercises, find the antiderivative $F\left(x\right)$ of each function $f\left(x\right).$
$f\left(x\right)=5{x}^{4}+4{x}^{5}$
$f\left(x\right)=x+12{x}^{2}$
$F\left(x\right)=\frac{1}{2}{x}^{2}+4{x}^{3}+C$
$f\left(x\right)=\frac{1}{\sqrt{x}}$
$f\left(x\right)={\left(\sqrt{x}\right)}^{3}$
$F\left(x\right)=\frac{2}{5}{\left(\sqrt{x}\right)}^{5}+C$
$f\left(x\right)={x}^{1\text{/}3}+{\left(2x\right)}^{1\text{/}3}$
$f\left(x\right)=\frac{{x}^{1\text{/}3}}{{x}^{2\text{/}3}}$
$F\left(x\right)=\frac{3}{2}{x}^{2\text{/}3}+C$
$f\left(x\right)=2\phantom{\rule{0.1em}{0ex}}\text{sin}\left(x\right)+\text{sin}\left(2x\right)$
$f\left(x\right)={\text{sec}}^{2}\left(x\right)+1$
$F\left(x\right)=x+\text{tan}\left(x\right)+C$
$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x$
$f\left(x\right)={\text{sin}}^{2}\left(x\right)\text{cos}\left(x\right)$
$F\left(x\right)=\frac{1}{3}{\text{sin}}^{3}\left(x\right)+C$
$f\left(x\right)=0$
$f\left(x\right)=\frac{1}{2}{\text{csc}}^{2}\left(x\right)+\frac{1}{{x}^{2}}$
$F\left(x\right)=-\frac{1}{2}\phantom{\rule{0.1em}{0ex}}\text{cot}\left(x\right)-\frac{1}{x}+C$
$f\left(x\right)=\text{csc}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}x+3x$
$f\left(x\right)=4\phantom{\rule{0.1em}{0ex}}\text{csc}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}x-\text{sec}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}x$
$F\left(x\right)=\text{\u2212}\text{sec}\phantom{\rule{0.1em}{0ex}}x-4\phantom{\rule{0.1em}{0ex}}\text{csc}\phantom{\rule{0.1em}{0ex}}x+C$
$f\left(x\right)=8\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}x\left(\text{sec}\phantom{\rule{0.1em}{0ex}}x-4\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)$
$f\left(x\right)=\frac{1}{2}{e}^{\mathrm{-4}x}+\text{sin}\phantom{\rule{0.1em}{0ex}}x$
$F\left(x\right)=-\frac{1}{8}{e}^{\mathrm{-4}x}-\text{cos}\phantom{\rule{0.1em}{0ex}}x+C$
For the following exercises, evaluate the integral.
$\int \left(\mathrm{-1}\right)dx$
$\int \text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx$
$\text{\u2212}\text{cos}\phantom{\rule{0.1em}{0ex}}x+C$
$\int \left(4x+\sqrt{x}\right)dx$
$\int \left(\text{sec}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}x+4x\right)dx$
$\int \left(4\sqrt{x}+\sqrt[4]{x}\right)dx$
$\frac{8}{3}{x}^{3\text{/}2}+\frac{4}{5}{x}^{5\text{/}4}+C$
$\int \left({x}^{\mathrm{-1}\text{/}3}-{x}^{2\text{/}3}\right)}dx$
$\int \frac{14{x}^{3}+2x+1}{{x}^{3}}dx$
$14x-\frac{2}{x}-\frac{1}{2{x}^{2}}+C$
$\int \left({e}^{x}+{e}^{\text{\u2212}x}\right)}dx$
For the following exercises, solve the initial value problem.
${f}^{\prime}\left(x\right)={x}^{\mathrm{-3}},f\left(1\right)=1$
$f\left(x\right)=-\frac{1}{2{x}^{2}}+\frac{3}{2}$
${f}^{\prime}\left(x\right)=\sqrt{x}+{x}^{2},f\left(0\right)=2$
${f}^{\prime}\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x+{\text{sec}}^{2}\left(x\right),f\left(\frac{\pi}{4}\right)=2+\frac{\sqrt{2}}{2}$
$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x+\text{tan}\phantom{\rule{0.1em}{0ex}}x+1$
${f}^{\prime}\left(x\right)={x}^{3}-8{x}^{2}+16x+1,f\left(0\right)=0$
${f}^{\prime}\left(x\right)=\frac{2}{{x}^{2}}-\frac{{x}^{2}}{2},f\left(1\right)=0$
$f\left(x\right)=-\frac{1}{6}{x}^{3}-\frac{2}{x}+\frac{13}{6}$
For the following exercises, find two possible functions $f$ given the second- or third-order derivatives.
$f\text{\u2033}\left(x\right)={x}^{2}+2$
$f\text{\u2033}\left(x\right)={e}^{\text{\u2212}x}$
Answers may vary; one possible answer is $f\left(x\right)={e}^{\text{\u2212}x}$
$f\text{\u2033}\left(x\right)=1+x$
$f\text{\u2034}\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$
Answers may vary; one possible answer is $f\left(x\right)=\text{\u2212}\text{sin}\phantom{\rule{0.1em}{0ex}}x$
$f\text{\u2034}\left(x\right)=8{e}^{\mathrm{-2}x}-\text{sin}\phantom{\rule{0.1em}{0ex}}x$
A car is being driven at a rate of $40$ mph when the brakes are applied. The car decelerates at a constant rate of $10$ ft/sec ^{2} . How long before the car stops?
$5.867$ sec
In the preceding problem, calculate how far the car travels in the time it takes to stop.
You are merging onto the freeway, accelerating at a constant rate of $12$ ft/sec ^{2} . How long does it take you to reach merging speed at $60$ mph?
$7.333$ sec
Based on the previous problem, how far does the car travel to reach merging speed?
A car company wants to ensure its newest model can stop in $8$ sec when traveling at $75$ mph. If we assume constant deceleration, find the value of deceleration that accomplishes this.
$13.75$ ft/sec ^{2}
A car company wants to ensure its newest model can stop in less than $450$ ft when traveling at $60$ mph. If we assume constant deceleration, find the value of deceleration that accomplishes this.
For the following exercises, find the antiderivative of the function, assuming $F\left(0\right)=0.$
[T] $f\left(x\right)={x}^{2}+2$
$F\left(x\right)=\frac{1}{3}{x}^{3}+2x$
[T] $f\left(x\right)=4x-\sqrt{x}$
[T] $f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x+2x$
$F\left(x\right)={x}^{2}-\text{cos}\phantom{\rule{0.1em}{0ex}}x+1$
[T] $f\left(x\right)={e}^{x}$
[T] $f\left(x\right)=\frac{1}{{\left(x+1\right)}^{2}}$
$F\left(x\right)=-\frac{1}{\left(x+1\right)}+1$
[T] $f\left(x\right)={e}^{\mathrm{-2}x}+3{x}^{2}$
For the following exercises, determine whether the statement is true or false. Either prove it is true or find a counterexample if it is false.
If $f\left(x\right)$ is the antiderivative of $v\left(x\right),$ then $2f\left(x\right)$ is the antiderivative of $2v\left(x\right).$
True
If $f\left(x\right)$ is the antiderivative of $v\left(x\right),$ then $f\left(2x\right)$ is the antiderivative of $v\left(2x\right).$
If $f\left(x\right)$ is the antiderivative of $v\left(x\right),$ then $f\left(x\right)+1$ is the antiderivative of $v\left(x\right)+1.$
False
If $f\left(x\right)$ is the antiderivative of $v\left(x\right),$ then ${\left(f\left(x\right)\right)}^{2}$ is the antiderivative of ${\left(v\left(x\right)\right)}^{2}.$
True or False ? Justify your answer with a proof or a counterexample. Assume that $f\left(x\right)$ is continuous and differentiable unless stated otherwise.
If $f\left(\mathrm{-1}\right)=\mathrm{-6}$ and $f\left(1\right)=2,$ then there exists at least one point $x\in \left[\mathrm{-1},1\right]$ such that ${f}^{\prime}\left(x\right)=4.$
True, by Mean Value Theorem
If ${f}^{\prime}\left(c\right)=0,$ there is a maximum or minimum at $x=c.$
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