6.1 Areas between curves  (Page 4/4)

$y=x^2-3\text{and}y=1$

$y=x^2\text{and}y=3x+4$

$y=x^3$ and $y=x^2+x$

$y=\text{cos}\theta$ and $y=0.5,$ for $0\le \theta \le \pi$

$x=y^2\text{and}x=9$

$y=x\text{and}x=y^2$

$y=x^2\text{and}y=\text{−}{x}^2+18x$

$y=\frac{1}{x},y=\frac{1}{x^2},\text{and}x=3$

$y=\text{cos}x$ and $y={\text{cos}}^{2}x$ on $x=\left[\text{−}\pi ,\pi \right]$

$y=e^x,y=e^{2x-1},\text{and}x=0$

$y=e^x,y=e^{\text{−}x},x=\mathrm{-1}\text{and}x=1$

$y=e,y=e^x,\text{and}y=e^{\text{−}x}$

$y=\left|x\right|\text{and}y=x^2$

$y=\text{sin}\left(\pi x\right),y=2x,\text{and}x>0$

$y=12-x,y=\sqrt{x},\text{and}y=1$

$y=\text{sin}x$ and $y=\text{cos}x$ over $x=\left[\text{−}\pi ,\pi \right]$

$y=x^3\text{and}y=x^2-2x$ over $x=\left[\mathrm{-1},1\right]$

$y=x^2+9\text{and}y=10+2x$ over $x=\left[\mathrm{-1},3\right]$

$y=x^3+3x$ and $y=4x$

$x=y^3\text{and}x=3y-2$

$x=2y\text{and}x=y^3-y$

$x=\mathrm{-3}+y^2\text{and}x=y-y^2$

$y^2=x\text{and}x=y+2$

$x=\left|y\right|\text{and}2x=\text{−}{y}^2+2$

$x=\text{sin}y,x=\text{cos}(2y),y=\pi \text{/}2,\text{and}y=\text{−}\pi \text{/}2$

$x=y^4\text{and}x=y^5$

$y=x{e}^x,y=e^x,x=0,\text{and}x=1$

$y=x^6\text{and}y=x^4$

$x=y^3+2y^2+1\text{and}x=\text{−}{y}^2+1$

$y=\left|x\right|\text{and}y=x^2-1$

$y=4-3x\text{and}y=\frac{1}{x}$

$y=\text{sin}x,x=\text{−}\pi \text{/}6,x=\pi \text{/}6,\text{and}y={\text{cos}}^{3}x$

$y=x^2-3x+2\text{and}y=x^3-2x^2-x+2$

$y=2{\text{cos}}^3\left(3x\right),y=\mathrm{-1},x=\frac{\pi }{4},\text{and}x=-\frac{\pi }{4}$

$y+y^3=x\text{and}2y=x$

$y=\sqrt{1-x^2}\text{and}y=x^2-1$

$y={\text{cos}}^{\mathrm{-1}}x,y={\text{sin}}^{\mathrm{-1}}x,x=\mathrm{-1},\text{and}x=1$

[T] $x=e^y\text{and}y=x-2$

[T] $y=x^2\text{and}y=\sqrt{1-x^2}$

[T] $y=3x^2+8x+9\text{and}3y=x+24$

[T] $x=\sqrt{4-y^2}\text{and}{y}^2=1+x^2$

[T] $x^2=y^3\text{and}x=3y$

[T] $y={\text{sin}}^{3}x+2,y=\text{tan}x,x=\mathrm{-1.5},\text{and}x=1.5$

[T] $y=\sqrt{1-x^2}\text{and}{y}^2=x^2$

[T] $y=\sqrt{1-x^2}\text{and}y=x^2+2x+1$

[T] $x=4-y^2\text{and}x=1+3y+y^2$

[T] $y=\text{cos}x,y=e^x,x=\text{−}\pi ,\text{and}x=0$

The largest triangle with a base on the $x\text{-axis}$ that fits inside the upper half of the unit circle $y^2+x^2=1$ is given by $y=1+x$ and $y=1-x.$ See the following figure. What is the area inside the semicircle but outside the triangle?

A factory selling cell phones has a marginal cost function $C(x)=0.01x^2-3x+229,$ where $x$ represents the number of cell phones, and a marginal revenue function given by $R(x)=429-2x.$ Find the area between the graphs of these curves and $x=0.$ What does this area represent?

An amusement park has a marginal cost function $C(x)=1000e^{\text{−}x}+5,$ where $x$ represents the number of tickets sold, and a marginal revenue function given by $R(x)=60-0.1x.$ Find the total profit generated when selling $550$ tickets. Use a calculator to determine intersection points, if necessary, to two decimal places.

The tortoise versus the hare: The speed of the hare is given by the sinusoidal function $H(t)=1-\text{cos}\left(\left(\pi t\right)\text{/}2\right)$ whereas the speed of the tortoise is $T(t)=\left(1\text{/}2\right){\text{tan}}^{\mathrm{-1}}\left(t\text{/}4\right),$ where $t$ is time measured in hours and the speed is measured in miles per hour. Find the area between the curves from time $t=0$ to the first time after one hour when the tortoise and hare are traveling at the same speed. What does it represent? Use a calculator to determine the intersection points, if necessary, accurate to three decimal places.

The tortoise versus the hare: The speed of the hare is given by the sinusoidal function $H(t)=\left(1\text{/}2\right)-\left(1\text{/}2\right)\text{cos}\left(2\pi t\right)$ whereas the speed of the tortoise is $T(t)=\sqrt{t},$ where $t$ is time measured in hours and speed is measured in kilometers per hour. If the race is over in $1$ hour, who won the race and by how much? Use a calculator to determine the intersection points, if necessary, accurate to three decimal places.

$y=x^2+2x+1\text{and}y=\text{−}{x}^2-3x+4$

$y=x^4\text{and}x=y^5$

$x=y^2-2\text{and}x=2y$

Determine the equations for the sides of the square that touches the unit circle on all four sides, as seen in the following figure. Find the area between the perimeter of this square and the unit circle. Is there another way to solve this without using calculus?

Find the area between the perimeter of the unit circle and the triangle created from $y=2x+1,y=1-2x$ and $y=-\frac{3}{5},$ as seen in the following figure. Is there a way to solve this without using calculus?