4.10 Antiderivatives  (Page 6/10)

$f\left(x\right)=\frac{1}{x^2}+x$

$f\left(x\right)=e^x-3x^2+\text{sin}x$

$f\left(x\right)=e^x+3x-x^2$

$f\left(x\right)=x-1+4\text{sin}\left(2x\right)$

$f\left(x\right)=5x^4+4x^5$

$f\left(x\right)=x+12x^2$

$f\left(x\right)=\frac{1}{\sqrt{x}}$

$f\left(x\right)={\left(\sqrt{x}\right)}^3$

$f\left(x\right)=x^{1\text{/}3}+{\left(2x\right)}^{1\text{/}3}$

$f\left(x\right)=\frac{x^{1\text{/}3}}{x^{2\text{/}3}}$

$f\left(x\right)=2\text{sin}\left(x\right)+\text{sin}\left(2x\right)$

$f\left(x\right)={\text{sec}}^2\left(x\right)+1$

$f\left(x\right)=\text{sin}x\text{cos}x$

$f\left(x\right)={\text{sin}}^2\left(x\right)\text{cos}\left(x\right)$

$f\left(x\right)=0$

$f\left(x\right)=\frac{1}{2}{\text{csc}}^2\left(x\right)+\frac{1}{x^2}$

$f\left(x\right)=\text{csc}x\text{cot}x+3x$

$f\left(x\right)=4\text{csc}x\text{cot}x-\text{sec}x\text{tan}x$

$f\left(x\right)=8\text{sec}x\left(\text{sec}x-4\text{tan}x\right)$

$f\left(x\right)=\frac{1}{2}{e}^{\mathrm{-4}x}+\text{sin}x$

${\displaystyle \int \left(\mathrm{-1}\right)dx}$

${\displaystyle \int \text{sin}xdx}$

${\displaystyle \int \left(4x+\sqrt{x}\right)dx}$

${\displaystyle \int \frac{3x^2+2}{x^2}dx}$

${\displaystyle \int \left(\text{sec}x\text{tan}x+4x\right)dx}$

${\displaystyle \int \left(4\sqrt{x}+\sqrt[4]{x}\right)dx}$

${\displaystyle \int \left(x^{\mathrm{-1}\text{/}3}-x^{2\text{/}3}\right)}dx$

${\displaystyle \int \frac{14x^3+2x+1}{x^3}dx}$

${\displaystyle \int \left(e^x+e^{\text{−}x}\right)}dx$

$f^{\prime }\left(x\right)=x^{\mathrm{-3}},f\left(1\right)=1$

$f^{\prime }\left(x\right)=\sqrt{x}+x^2,f\left(0\right)=2$

$f^{\prime }\left(x\right)=\text{cos}x+{\text{sec}}^2\left(x\right),f\left(\frac{\pi }{4}\right)=2+\frac{\sqrt{2}}{2}$

$f^{\prime }\left(x\right)=x^3-8x^2+16x+1,f\left(0\right)=0$

$f^{\prime }\left(x\right)=\frac{2}{x^2}-\frac{x^2}{2},f\left(1\right)=0$

$f\text{″}\left(x\right)=x^2+2$

$f\text{″}\left(x\right)=e^{\text{−}x}$

$f\text{″}\left(x\right)=1+x$

$f\text{‴}\left(x\right)=\text{cos}x$

$f\text{‴}\left(x\right)=8e^{\mathrm{-2}x}-\text{sin}x$

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?

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?

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.

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.

[T] $f\left(x\right)=x^2+2$

[T] $f\left(x\right)=4x-\sqrt{x}$

[T] $f\left(x\right)=\text{sin}x+2x$

[T] $f\left(x\right)=e^x$

[T] $f\left(x\right)=\frac{1}{{\left(x+1\right)}^2}$

[T] $f\left(x\right)=e^{\mathrm{-2}x}+3x^2$

If $f\left(x\right)$ is the antiderivative of $v\left(x\right),$ then $2f\left(x\right)$ is the antiderivative of $2v\left(x\right).$

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 $v\left(x\right)+1.$

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 ${\left(v\left(x\right)\right)}^2.$

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 $f^{\prime }\left(x\right)=4.$

If $f^{\prime }\left(c\right)=0,$ there is a maximum or minimum at $x=c.$