【高校数学Ⅲ】6-1-1 面積｜問題集

求める面積\(S\)は
\(\displaystyle S=\int_1^e\left(\frac{1}{x}\right)dx\)
\(\displaystyle =[\log x]_1^e\)
\(\displaystyle =1\)

求める面積\(S\)は
\(\displaystyle S=\int_0^3\sqrt{x+1}dx\)
\(\displaystyle =\frac{2}{3}[(x+1)^{\frac{3}{2}}]_0^3\)
\(\displaystyle =\frac{2}{3}(8-1)\)
\(\displaystyle =\frac{14}{3}\)

\(y=e^x-e\)と\(x\)軸との交点は
\(e^x-e=0\)
\(e^x=e\)
\(x=1\)
\(0\leqq x\leqq1\)のとき、\(e^x-e\leqq0\)
\(1\leqq x\leqq2\)のとき、\(e^x-e\geqq0\)
よって、求める面積\(S\)は
\(\displaystyle S=-\int_0^1(e^x-e)dx+\int_1^2(e^x-e)dx\)
\(\displaystyle =[e^x-1]_0^1+[e^x-1]_1^2\)
\(=-(e-1)+(e^2-1)-(e-1)\)
\(=e^2-2e+1\)

\(y=x^2\)と\(y=\sqrt{x}\)との交点は
\(x^2=\sqrt{x}\)
\(x^3-x=0\)
\(x(x+1)(x-1)=0\)
\(x=0,1,-1\)
\(0\leqq x\leqq1\)のとき、\(x^2\leqq\sqrt{x}\)
よって、求める面積\(S\)は
\(\displaystyle S=\int_0^1(\sqrt{x}-x^2)dx\)
\(\displaystyle =\left[\frac{2}{3}x^{\frac{3}{2}}-\frac{1}{3}x^3\right]_0^1\)
\(\displaystyle =\frac{2}{3}-\frac{1}{3}\)
\(\displaystyle =\frac{1}{3}\)

\(x+4y=5\)と\(xy=1\)との交点は
\(\displaystyle x+\frac{4}{x}=5\)
\(x^2-5x+4=0\)
\((x-1)(x-4)=0\)
\(x=1,4\)
\(1\leqq x\leqq4\)のとき、\(\displaystyle -\frac{1}{4}x+\frac{5}{4}\geqq\frac{1}{x}\)
よって、求める面積\(S\)は
\(\displaystyle S=\int_1^4(-\frac{1}{4}x+\frac{5}{4}-\frac{1}{x})dx\)
\(\displaystyle =\left[-\frac{1}{8}x^2+\frac{5}{4}x-\log x\right]_1^4\)
\(\displaystyle =-2+5-\log4+\frac{1}{8}-\frac{5}{4}+\log1\)
\(\displaystyle =\frac{15}{8}-2\log2\)

\(y=\sqrt{2-x^2}\)と\(y=x^2\)との交点は
\(\sqrt{2-x^2}=x^2\)
\(x^4+x^2-2=0\)
\((x^2+2)(x-1)(x+1)=0\)
\(x=-2,-1,1\)
\(-1\leqq x\leqq1\)のとき、\(\sqrt{2-x^2}\geqq x^2\)
よって、求める面積\(S\)は
\(\displaystyle S=\int_{-1}^1(\sqrt{2-x^2}-x^2)dx\)
\(x=\sqrt{2}\sin\theta\)とおくと、\(dx=\sqrt{2}\cos\theta d\theta\)
\(\displaystyle =2\int_0^{\frac{\pi}{4}}\sqrt{2}\cos\theta・\sqrt{2}\cos\theta d\theta-2\int_0^1 x^2dx\)
\(\displaystyle =4\int_0^{\frac{\pi}{4}}\frac{1+\cos2\theta}{2}d\theta-2\left[\frac{1}{3}x^3\right]_0^1\)
\(\displaystyle =2\left[\theta+\frac{1}{2}\sin2\theta\right]_0^{\frac{\pi}{4}}-\frac{2}{3}\)
\(\displaystyle =2\left(\frac{\pi}{4}-\frac{1}{2}\right)-\frac{2}{3}\)
\(\displaystyle =\frac{\pi}{2}+\frac{1}{3}\)

求める面積\(S\)は
\(\displaystyle S=\int_0^2 (y^2+1)dy\)
\(\displaystyle =\left[\frac{y^3}{3}+y\right]_0^2\)
\(\displaystyle =\frac{14}{3}\)

\(x=y^2-1\)と\(x=y+5\)との交点は
\(y^2-1=y+5\)
\(y^2-y-6=0\)
\((y-3)(y+2)=0\)
\(y=-2,3\)
よって、求める面積\(S\)は
\(\displaystyle S=\int_{-2}^3 -(y-3)(y-2)dy\)
\(\displaystyle =\frac{(3+2)^3}{6}\)
\(\displaystyle =\frac{125}{6}\)

求める面積\(S\)は
\(\displaystyle S=\int_1^4 \left(\sqrt{y}+\frac{1}{\sqrt{y}}\right)dy\)
\(\displaystyle =\left[\frac{2}{3}y^{\frac{3}{2}}+2y^{\frac{1}{2}}\right]_1^4\)
\(\displaystyle =\frac{2}{3}・4・2+2・2-\left(\frac{2}{3}+2\right)\) \(\displaystyle =\frac{20}{3}\)

\(4x^2+2y^2=1\)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。
\(4x^2+2y^2=1\)より
\(\displaystyle y^2=\frac{1}{2}(1-4x^2)\)
\(\displaystyle y=\frac{1}{\sqrt{2}}\sqrt{1-4x^2}\)
求める面積\(S\)は
\(\displaystyle S=\int_0^{\frac{1}{2}} \frac{1}{\sqrt{2}}\sqrt{1-4x^2}dx\)
\(\displaystyle =\frac{1}{\sqrt{2}}・\frac{1}{2}・\left(\frac{1}{2}\right)^2\pi\)
\(\displaystyle =\frac{\sqrt{2}}{16}\pi\)
よって、
\(\displaystyle 4・\frac{\sqrt{2}}{16}\pi=\frac{\sqrt{2}}{4}\pi\)

求める面積\(S\)は
\(\displaystyle S=\int_0^1 (1-\sqrt{x})^2dx\)
\(\displaystyle =\int_0^1 (1-2\sqrt{x}+x)dx\)
\(\displaystyle =\left[x-\frac{4}{3}x^{\frac{3}{2}}+\frac{1}{2}x^2\right]_0^1\)
\(\displaystyle =1-\frac{4}{3}+\frac{1}{2}\)
\(\displaystyle =\frac{1}{6}\)

\(y=2\sin\theta\)より、\(dx=-3\sin\theta d\theta\)
求める面積\(S\)は
\(\displaystyle S=\int_{-3}^3 ydx\)
\(\displaystyle =\int_\pi^0 2\sin\theta)・(-3\sin\theta)d\theta\)
\(\displaystyle =6\int_0^{\pi} \sin^2\theta d\theta\)
\(\displaystyle =6\int_0^{\pi} \frac{1-\cos2\theta}{2}d\theta\)
\(\displaystyle =3\left[\theta-\frac{1}{2}\sin2\theta\right]_0^{\pi}\)
\(\displaystyle =3\pi\)