§6.1 Using definite integrals to find area and length (continued)

Summary

Between $x = \frac{\pi}{4}$ and $x = \frac{3\pi}{4}$, the area between $y = \sin(x)$ and $y = \cos(x)$ equals

For $f(x) = 1-x^2$ and $g(x) = ax^2 - a$ with $a \gt 0$, the intersection points of the two graphs are

With $f(x) = 1-x^2$ above $g(x) = ax^2-a$ on $(-1,1)$, setting the area between the curves equal to $2$ determines $a$. The solution is

For the region between $x = y(y-2)$ and $x = -(y-1)(y-3)$, which lies between $y = \frac{3-\sqrt{3}}{2}$ and $y = \frac{3+\sqrt{3}}{2}$, the area is $\sqrt{3}$. Report this as a decimal, rounded to three places.

For the region between $x = y^2-y-2$ and $x = \frac{y+1}{2}$ on $-1 \le y \le \frac{5}{2}$, the area is $\frac{343}{48}$. Report this as a decimal, rounded to three places.

The curves $x = y(y-2)$ and $x = -(y-1)(y-3)$ intersect at $y = \frac{3 \pm \sqrt{3}}{2}$.