§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 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.

For the region between $y = mx$ and $y = x^2-1$ with $m \gt 0$, the two curves intersect at the $x$-values

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

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 $f(x) = 2-x^2$ on the interval $[0,\sqrt{2}]$, compute the average value $r = \frac{1}{\sqrt{2}}\int_0^{\sqrt{2}}(2-x^2)\,dx$ as a decimal, rounded to three places. The exact value is $\frac{4}{3}$.

For any continuous function on an interval, the area above the line $y = AV$ and below the graph equals the area below the line and above the graph.