【高校数学Ⅲ】6-1-1 面積｜要点まとめ

\(\displaystyle y=\frac{x}{x^2+1}\)
\(\displaystyle y'=\frac{-(x+1)(x-1)}{(x^2+1)^2}\)
増減表にまとめると、

増減表

\(x\)	\(\cdots\)	\(-1\)	\(\cdots\)	\(1\)	\(\cdots\)
\(f'(x)\)	\(-\)	\(0\)	\(+\)	\(0\)	\(-\)
\(f(x)\)	\(\searrow\)	\(\displaystyle -\frac{1}{2}\)	\(\nearrow\)	\(\displaystyle \frac{1}{2}\)	\(\searrow\)

\(\displaystyle y=\frac{x}{x^2+1}\)と\(\displaystyle y=\frac{1}{2}x\)との交点は
\(\displaystyle \frac{x}{x^2+1}=\frac{1}{2}x\)
\(x(x^2+1)-2x=0\)
\(x(x-1)(x+1)=0\)
\(x=-1,0,1\)
\(-1\leqq x\leqq 0\)のとき、\(\displaystyle \frac{x}{x^2+1}\leqq\frac{1}{2}x\)
\(0\leqq x\leqq 1\)のとき、\(\displaystyle \frac{1}{2}x\leqq\frac{x}{x^2+1}\)
よって、求める面積\(S\)は
\(\displaystyle S=\int_{-1}^0\left(\frac{1}{2}x-\frac{x}{x^2+1}\right)dx\)
\(\displaystyle \ \ \ +\int_0^1\left(\frac{x}{x^2+1}-\frac{1}{2}x\right)dx\)
\(\displaystyle =\left[\frac{1}{4}x^2-\frac{1}{2}\log|x^2+1|\right]_{-1}^0\)
\(\displaystyle \ \ \ +\left[\frac{1}{2}\log|x^2+1|-\frac{1}{4}x^2\right]_0^1\)
\(\displaystyle =-\left(\frac{1}{4}-\frac{1}{2}\log2\right)+\frac{1}{2}\log2-\frac{1}{4}\)
\(\displaystyle =\log2-\frac{1}{2}\)

\(y=\log x\)を\(x\)について解くと
\(x=e^y\)
よって、求める面積\(S\)は
\(\displaystyle S=\int_1^2 e^ydy\)
\(\displaystyle =[e^y]_1^2\)
\(=e^2-e\)

\(\displaystyle \frac{x^2}{9}+\frac{y^2}{16}=1\)は\(x\)軸、\(y\)軸に関して対称なので、\(x\geqq0,y\geqq0\)の部分の面積を求めて\(4\)倍する。
\(\displaystyle \frac{x^2}{9}+\frac{y^2}{16}=1\)より
\(\displaystyle y^2=16-\frac{16}{9}x^2\)
\(\displaystyle y=\frac{4}{3}\sqrt{9-x^2}\)
求める面積\(S\)は
\(\displaystyle S=\int_0^3 \frac{4}{3}\sqrt{9-x^2}dx\)
\(x=3\sin\theta\)とおくと、\(dx=3\cos\theta d\theta\)なので、
\(\displaystyle =\int_0^{\frac{\pi}{2}} \frac{4}{3}・3\cos\theta・3\cos\theta d\theta\)
\(\displaystyle =12\int_0^{\frac{\pi}{2}} \cos^2\theta d\theta\)
\(\displaystyle =12\int_0^{\frac{\pi}{2}} \frac{1+\cos2\theta}{2} d\theta\)
\(\displaystyle =[6\theta+3\sin2\theta]_0^{\frac{\pi}{2}}\)
\(=3\pi\)
よって、
\(4・3\pi=12\pi\)

\(y=1-\cos\theta\)より、\(dx=(1\cos\theta) d\theta\)
求める面積\(S\)は
\(\displaystyle S=\int_0^{2\pi} ydx\)
\(\displaystyle =\int_0^{2\pi} (1-\cos\theta)・(1-\cos\theta)d\theta\)
\(\displaystyle =\int_0^{2\pi} (1-2\cos\theta+\cos^2\theta)d\theta\)
\(\displaystyle =\int_0^{2\pi} \left(1-2\cos\theta+\frac{1+\cos2\theta}{2}\right)d\theta\)
\(\displaystyle =\left[\theta-2\sin\theta+\frac{1}{2}\theta+\frac{1}{4}\sin2\theta\right]_0^{2\pi}\)
\(=3\pi\)