【高校数学Ⅱ】6-3-3 定積分と面積｜問題集

\(y=-2x(x-2)\)
\(0\leqq x\leqq2\)において、\(y\geqq0\)なので、
\(\displaystyle S=\int_0^2(-2x^2+4x)dx\)
\(\displaystyle \ \ =\left[-\frac{2}{3}x^3+2x^2\right]_0^2\)
\(\displaystyle \ \ =\frac{8}{3}\)

\(y=-(x+1)(x-2)\)
\(-1\leqq x\leqq2\)において、\(y\geqq0\)なので、
\(\displaystyle S=\int_{-1}^2(-x^2+x+2)dx\)
\(\displaystyle \ \ =\left[-\frac{1}{3}x^3+\frac{1}{2}x^2+2x\right]_{-1}^2\)
\(\displaystyle \ \ =\frac{10}{3}+\frac{7}{6}\)
\(\displaystyle \ \ =\frac{9}{2}\)

\(y=(x-1)(x-4)\)
\(1\leqq x\leqq4\)において、\(y\leqq0\)なので、
\(\displaystyle S=-\int_1^4(x^2-5x+4)dx\)
\(\displaystyle \ \ =\left[-\frac{1}{3}x^3+\frac{5}{2}x^2-4x\right]_1^4\)
\(\displaystyle \ \ =\frac{8}{3}+\frac{11}{6}\)
\(\displaystyle \ \ =\frac{9}{2}\)

\(y=-(x+2)(x-2)\)
\(-2\leqq x\leqq2\)において、\(y\geqq0\)なので、
\(\displaystyle S=\int_{-2}^2(-x^2+4)dx\)
\(\displaystyle \ \ =\left[-\frac{1}{3}x^3+4x\right]_{-2}^2\)
\(\displaystyle \ \ =\frac{16}{3}+\frac{16}{3}\)
\(\displaystyle \ \ =\frac{32}{3}\)

\(y=(x-1)(x-2)\)
\(1\leqq x\leqq2\)において、\(y\leqq0\)なので、
\(\displaystyle S=-\int_1^2(x^2-3x+2)dx\)
\(\displaystyle \ \ =\left[-\frac{1}{3}x^3-\frac{3}{2}x^2+2x\right]_1^2\)
\(\displaystyle \ \ =-\frac{2}{3}+\frac{5}{6}\)
\(\displaystyle \ \ =\frac{1}{6}\)

\(-1\leqq x\leqq2\)において、\(y\geqq0\)なので、
\(\displaystyle S=\int_{-1}^2(x^2+2)dx\)
\(\displaystyle \ \ =\left[\frac{1}{3}x^3+2x\right]_{-1}^2\)
\(\displaystyle \ \ =\frac{20}{3}+\frac{7}{3}\)
\(\ \ =9\)

\(-1\leqq x\leqq1\)において、\(y\geqq0\)なので、
\(\displaystyle S=\int_{-1}^1(x^2+2x+2)dx\)
\(\displaystyle \ \ =\left[\frac{1}{3}x^3+x^2+2x\right]_{-1}^1\)
\(\displaystyle \ \ =\frac{10}{3}+\frac{4}{3}\)
\(\displaystyle \ \ =\frac{14}{3}\)

\(1\leqq x\leqq2\)において、\(y\geqq0\)なので、
\(\displaystyle S=\int_1^2(2x^2+2x)dx\)
\(\displaystyle \ \ =\left[\frac{2}{3}x^3+x^2\right]_1^2\)
\(\displaystyle \ \ =\frac{28}{3}-\frac{5}{3}\)
\(\displaystyle \ \ =\frac{23}{3}\)

\(y=(x+1)(x-1)\)
\(0\leqq x\leqq1\)において、\(y\leqq0\)
\(1\leqq x\leqq3\)において、\(y\geqq0\)なので、
\(\displaystyle S=-\int_0^1(x^2-1)dx+\int_1^3(x^2-1)dx\)
\(\displaystyle \ \ =\left[-\frac{1}{3}x^3+x\right]_0^1+\left[\frac{1}{3}x^3-x\right]_1^3\)
\(\displaystyle \ \ =\left(\frac{2}{3}-0\right)+\left(6+\frac{2}{3}\right)\)
\(\displaystyle \ \ =\frac{22}{3}\)

\(y=x^3-3x^2\)
\(0\leqq x\leqq3\)において、\(y\leqq0\)なので、
\(\displaystyle S=-\int_0^3(x^3-3x^2)dx\)
\(\displaystyle \ \ =\left[-\frac{1}{4}x^4+x^3\right]_0^3\)
\(\displaystyle \ \ =-\frac{81}{4}+27\)
\(\displaystyle \ \ =\frac{27}{4}\)

\(y=x(x+1)(x-2)\)
\(-1\leqq x\leqq0\)において、\(y\geqq0\)
\(0\leqq x\leqq2\)において、\(y\leqq0\)なので、
\(\displaystyle S=\int_{-1}^0(x^3-x^2-2x)dx-\int_0^2(x^3-x^2-2x)dx\)
\(\displaystyle \ \ =\left[\frac{1}{4}x^4-\frac{1}{3}x^3-x^2\right]_{-1}^0+\left[-\frac{1}{4}x^4+\frac{1}{3}x^3+x^2\right]_0^2\)
\(\displaystyle \ \ =\left(0+\frac{5}{12}\right)+\left(\frac{8}{3}-0\right)\)
\(\displaystyle \ \ =\frac{37}{12}\)

\(y=(x+1)(x-1)(x-3)\)
\(-1\leqq x\leqq1\)において、\(y\geqq0\)
\(1\leqq x\leqq3\)において、\(y\leqq0\)なので、
\(\displaystyle S=\int_{-1}^1(x^3-3x^2-x+3)dx\)
\(\displaystyle \ \ \ -\int_1^3(x^3-3x^2-x+3)dx\)
\(\displaystyle \ \ =\left[\frac{1}{4}x^4-x^3-\frac{1}{2}x^2+3x\right]_{-1}^1\)
\(\displaystyle \ \ \ +\left[-\frac{1}{4}x^4+x^3+\frac{1}{2}x^2-3x\right]_1^3\)
\(\ \ =4+4\)
\(\ \ =8\)

\(y=(x+3)(x-1)(x-2)\)
\(-3\leqq x\leqq1\)において、\(y\geqq0\)
\(1\leqq x\leqq2\)において、\(y\leqq0\)なので、
\(\displaystyle S=\int_{-3}^1(x^3-7x+6)dx-\int_1^2(x^3-7x+6)dx\)
\(\displaystyle \ \ =\left[\frac{1}{4}x^4-\frac{7}{2}x^2+6x\right]_{-3}^1+\left[-\frac{1}{4}x^4+\frac{7}{2}x^2-6x\right]_1^2\)
\(\displaystyle \ \ =\frac{131}{4}\)

\(x\)軸の交点を求める。
\(-2x^2+4=-2x\)
\(x^2-x-2=0\)
\((x+1)(x-2)=0\)
\(x=-1,2\)
\(-1\leqq x\leqq2\)において、\(y=-2x^2+4\)が上なので、
\(\displaystyle S=\int_{-1}^2\{(-2x^2+4)-(-2x)\}dx\)
\(\displaystyle \ \ =\int_{-1}^2(-2x^2+2x+4)dx\)
\(\displaystyle \ \ =\left[-\frac{2}{3}x^3+x^2+4x\right]_{-1}^2\)
\(\displaystyle \ \ =\frac{20}{3}-\left(-\frac{7}{3}\right)\)
\(\ \ =9\)

\(x\)軸の交点を求める。
\(x^2-2x=x\)
\(x^2-3x=0\)
\(x(x-3)=0\)
\(x=0,3\)
\(0\leqq x\leqq3\)において、\(y=x\)が上なので、
\(\displaystyle S=\int_0^3\{x-(x^2-2x)\}dx\)
\(\displaystyle \ \ =\int_0^3(-x^2+3x)dx\)
\(\displaystyle \ \ =\left[-\frac{1}{3}x^3+\frac{3}{2}x^2\right]_0^3\)
\(\displaystyle \ \ =-9+\frac{27}{2}\)
\(\displaystyle \ \ =\frac{9}{2}\)

\(x\)軸の交点を求める。
\(x^2+2x+3=-2x\)
\(x^2+4x+3=0\)
\((x+1)(x+3)=0\)
\(x=-1,-3\)
\(-3\leqq x\leqq-1\)において、\(y=-2x\)が上なので、
\(\displaystyle S=\int_{-3}^{-1}\{-2x-(x^2+2x+3)\}dx\)
\(\displaystyle \ \ =\int_{-3}^{-1}(-x^2-4x-3)dx\)
\(\displaystyle \ \ =\left[-\frac{1}{3}x^3-2x^2-3x\right]_{-3}^{-1}\)
\(\displaystyle \ \ =\frac{4}{3}\)

\(x\)軸の交点を求める。
\(x^2+x-2=-2x^2+4x+4\)
\(x^2-x-2=0\)
\((x+1)(x-2)=0\)
\(x=-1,2\)
\(-1\leqq x\leqq2\)において、\(y=-2x^2+4x+4\)が上なので、
\(\displaystyle S=\int_{-1}^2\{(-2x^2+4x+4)-(x^2+x-2)\}dx\)
\(\displaystyle \ \ =\int_{-1}^2(-3x^2+3x+6)dx\)
\(\displaystyle \ \ =\left[-x^3+\frac{3}{2}x^2+6x\right]_{-1}^2\)
\(\displaystyle \ \ =10-\left(-\frac{7}{2}\right)\)
\(\displaystyle \ \ =\frac{27}{2}\)

\(x\)軸の交点を求める。
\(x^2+x+1=2x^2-3x+1\)
\(x^2-4x=0\)
\(x(x-4)=0\)
\(x=0,4\)
\(0\leqq x\leqq4\)において、\(y=x^2+x+1\)が上なので、
\(\displaystyle S=\int_0^4\{(x^2+x+1)-(2x^2-3x+1)\}dx\)
\(\displaystyle \ \ =\int_0^4(-x^2+4x)dx\)
\(\displaystyle \ \ =\left[-\frac{1}{3}x^3+2x^2\right]_0^4\)
\(\displaystyle \ \ =-\frac{64}{3}+32\)
\(\displaystyle \ \ =\frac{32}{3}\)

\(x\)軸の交点を求める。
\(x^2-6=-x^2+2x+6\)
\(x^2-x-6=0\)
\((x+2)(x-3)=0\)
\(x=-2,3\)
\(-2\leqq x\leqq3\)において、\(y=-x^2+2x+6\)が上なので、
\(\displaystyle S=\int_{-2}^3\{(-x^2+2x+6)-(x^2-6)\}dx\)
\(\displaystyle \ \ =\int_{-2}^3(-2x^2+2x+12)dx\)
\(\displaystyle \ \ =\left[-\frac{2}{3}x^3+x^2+12x\right]_{-2}^3\)
\(\displaystyle \ \ =27+\frac{44}{3}\)
\(\displaystyle \ \ =\frac{125}{3}\)

\(x\)軸の交点を求める。
\((x+1)^2=-x^2+5\)
\(x^2+x-2=0\)
\((x+2)(x-1)=0\)
\(x=-2,1\)
\(-2\leqq x\leqq1\)において、\(y=-x^2+5\)が上なので、
\(\displaystyle S=\int_{-2}^1\{(-x^2+5)-(x+1)^2\}dx\)
\(\displaystyle \ \ =\int_{-2}^1(-2x^2-2x+4)dx\)
\(\displaystyle \ \ =\left[-\frac{2}{3}x^3-x^2+4x\right]_{-2}^1\)
\(\displaystyle \ \ =\frac{7}{3}+\frac{20}{3}\)
\(\ \ =9\)

\(x\)軸の交点を求める。
\(x^2-5x=-x^2+4x\)
\(2x^2-9x=0\)
\(x(2x-9)=0\)
\(\displaystyle x=0,\frac{9}{2}\)
\(1\leqq x\leqq2\)において、\(y=-x^2+4x\)が上なので、
\(\displaystyle S=\int_1^2\{(-x^2+4x)-(x^2-5x)\}dx\)
\(\displaystyle \ \ =\int_1^2(-2x^2+9x)dx\)
\(\displaystyle \ \ =\left[-\frac{2}{3}x^3+\frac{9}{2}x^2\right]_1^2\)
\(\displaystyle \ \ =\frac{38}{3}-\frac{23}{6}\)
\(\displaystyle \ \ =\frac{53}{6}\)

\(x\)軸の交点を求める。
\(x^2-2x+4=2x^2-4x+3\)
\(x^2-2x-1=0\)
\((x-1)^2=0\)
\(x=1\)
\(1\leqq x\leqq2\)において、\(y=x^2-2x+4\)が上なので、
\(\displaystyle S=\int_1^2\{(x^2-2x+4)-(2x^2-4x+3)\}dx\)
\(\displaystyle \ \ =\int_1^2(-x^2+2x+1)dx\)
\(\displaystyle \ \ =\left[-\frac{1}{3}x^3+x^2+x\right]_1^2\)
\(\displaystyle \ \ =\frac{10}{3}-\frac{5}{3}\)
\(\displaystyle \ \ =\frac{5}{3}\)

\(x\geqq1\)において、\(y=x-1\)
\(x<1\)において、\(y=-x+1\)なので、
\(\displaystyle S=\int_1^2(x-1)dx+\int_{-2}^1(-x+1)dx\)
\(\displaystyle \ \ =\left[\frac{1}{2}x^2-x\right]_1^2+\left[-\frac{1}{2}x^2+x\right]_{-2}^1\)
\(\displaystyle \ \ =\left(0+\frac{1}{2}\right)+\left(\frac{1}{2}+4\right)\)
\(\ \ =5\)

\(-1\leqq x\leqq0\)において、\(y=x^2-3x\)
\(0\leqq x\leqq3\)において、\(y=-x^2+3x\)
\(3\leqq x\leqq5\)において、\(y=x^2-3x\)なので、
\(\displaystyle S=\int_{-1}^0(x^2-3x)dx+\int_0^3(-x^2+3x)dx\)
\(\displaystyle \ \ \ +\int_3^5(x^2-3x)dx\)
\(\displaystyle \ \ =\left[\frac{1}{3}x^3-\frac{3}{2}x^2\right]_{-1}^0+\left[-\frac{1}{3}x^3+\frac{3}{2}x^2\right]_0^3\)
\(\displaystyle \ \ \ +\left[\frac{1}{3}x^3-\frac{3}{2}x^2\right]_3^5\)
\(\displaystyle \ \ =\left(0+\frac{11}{6}\right)+\left(\frac{9}{2}-0\right)+\left(\frac{25}{6}+\frac{9}{2}\right)\)
\(\ \ =15\)

\(f'(x)=3x^2+2x-2\)
\(f'(1)=3\)
よって、
\(y-0=3(x-1)\)
\(y=3x-3\)

\(x\)軸の交点を求める。
\(x^3+x^2-2x=3x-3\)
\(x^3+x^2-5x+3=0\)
\((x+3)(x-1)^2=0\)
\(x=-3,1\)
\(-3\leqq x\leqq-1\)において、\(y=x^3+x^2-2x\)が上なので、
\(\displaystyle S=\int_{-3}^1\{(x^3+x^2-2x)-(3x-3)\}dx\)
\(\displaystyle \ \ =\int_{-3}^1(x^3+x^2-5x+3)dx\)
\(\displaystyle \ \ =\left[\frac{1}{4}x^4+\frac{1}{3}x^3-\frac{5}{2}x^2+3x\right]_{-3}^1\)
\(\displaystyle \ \ =\frac{64}{3}\)

導関数は\(y'=2x+1\)
接点の座標を\((a,a^2+a)\)とすると、
接線の傾きは\(2a+1\)となる。
接線方程式は
\(y-(a^2+a)=(2a+1)(x-a)\)
\(y=(2a+1)x-a^2\)
この直線が\((0,-1)\)を通るので、
\(-1=-a^2\)
\(a^2-1=0\)
\((a+1)(a-1)=0\)
\(a=-1,1\)
よって、接線方程式は
\(y=-x-1,y=3x-1\)

\(-1\leqq x\leqq0\)において、\(y=x^2+x,y=-x-1\)
\(0\leqq x\leqq1\)において、\(y=x^2+x,y=3x-1\)なので、
\(\displaystyle S=\int_{-1}^0\{(x^2+x)-(-x-1)\}dx\)
\(\displaystyle \ \ \ +\int_0^1\{(x^2+x)-(3x-1)\}dx\)
\(\displaystyle \ \ =\left[\frac{1}{3}x^3+x^2+x\right]_{-1}^0+\left[\frac{1}{3}x^3-x^2+x\right]_0^1\)
\(\displaystyle \ \ =\left(0+\frac{1}{3}\right)+\left(\frac{1}{3}-0\right)\)
\(\displaystyle \ \ =\frac{2}{3}\)