1.次の定積分を求めなさい。
(1)\(\displaystyle \int_0^2 5dx\)
\(=[5x]_0^2\)
\(=10\)
(2)\(\displaystyle \int_0^2(x^2+4x-5)dx\)
\(\displaystyle =\left[\frac{1}{3}x^3+2x^2-5x\right]_0^2\)
\(\displaystyle =\frac{8}{3}-2\)
\(\displaystyle =\frac{2}{3}\)
(3)\(\displaystyle \int_0^2(-4x+1)dx\)
\(=[-2x^2+x]_0^2\)
\(=-8+2\)
\(=-6\)
(4)\(\displaystyle \int_{-1}^1(x^2-3)dx\)
\(\displaystyle =\left[\frac{1}{3}x^3-3x\right]_{-1}^1\)
\(\displaystyle =-\frac{8}{3}-\frac{8}{3}\)
\(\displaystyle =-\frac{16}{3}\)
(5)\(\displaystyle \int_2^3(x-2)(x-3)dx\)
\(\displaystyle =\int_2^3(x^2-5x+6)dx\)
\(\displaystyle =\left[\frac{1}{3}x^3-\frac{5}{2}x^2+6x\right]_2^3\)
\(\displaystyle =\frac{9}{2}-\frac{14}{3}\)
\(\displaystyle =-\frac{1}{6}\)
(6)\(\displaystyle \int_{-2}^2 x(x+2)^2dx\)
\(\displaystyle =\int_{-2}^2(x^3+4x^2+4x)dx\)
\(\displaystyle =\int_{-2}^2 4x^2dx\)
\(\displaystyle =\int_0^2 8x^2dx\)
\(\displaystyle =\left[\frac{8}{3}x^3\right]_0^2\)
\(\displaystyle =\frac{64}{3}\)
(7)\(\displaystyle \int_{-1}^3(x+3)(x-3)dx\)
\(\displaystyle =\int_{-1}^3(x^2-9)dx\)
\(\displaystyle =\left[\frac{1}{3}x^3-9x\right]_{-1}^3\)
\(\displaystyle =-18-\frac{26}{3}\)
\(\displaystyle =-\frac{80}{3}\)
(8)\(\displaystyle \int_0^2(2x+1)(4x^2-2x+1)dx\)
\(\displaystyle =\int_0^2(8x^3+1)dx\)
\(=[2x^4+x]_0^2\)
\(=34\)
(9)\(\displaystyle \int_{-1}^1(x+2)^2dx-\int_{-1}^1(x-2)^2dx\)
\(\displaystyle =\int_{-1}^1\{(x+2)^2-(x-2)^2\}dx\)
\(\displaystyle =\int_{-1}^1 8xdx\)
\(=[4x^2]_{-1}^1\)
\(=4-4\)
\(=0\)
(10)\(\displaystyle \int_{-1}^2(x^2-x)dx-\int_{3}^2(x^2-x)dx\)
\(\displaystyle =\int_{-1}^2(x^2-x)dx+\int_{2}^3(x^2-x)dx\)
\(\displaystyle =\int_{-1}^3(x^2-x)dx\)
\(\displaystyle =\left[\frac{1}{3}x^3-\frac{1}{2}x^2\right]_{-1}^3\)
\(\displaystyle =\frac{9}{2}+\frac{5}{6}\)
\(\displaystyle =\frac{16}{3}\)
(11)\(\displaystyle 2\int_{-2}^1(x^2+3x+3)dx\)
\(\displaystyle \ \ \ +3\int_1^{-2}(x+1)(x+2)dx\)
\(\displaystyle =\int_{-2}^1(2x^2+6x+6)dx-\int_{-2}^1(3x^2+9x+6)dx\)
\(\displaystyle =\int_{-2}^1(-x^2-3x)dx\)
\(\displaystyle =\left[-\frac{1}{3}x^3-\frac{3}{2}x^2\right]_{-2}^1\)
\(\displaystyle =-\frac{11}{6}+\frac{10}{3}\)
\(\displaystyle =\frac{3}{2}\)
(12)\(\displaystyle \int_1^3 x^2(x-4)dx+4\int_1^3 x(x-1)dx\)
\(\displaystyle \ \ \ -\int_2^3 x(x+2)(x-2)dx\)
\(\displaystyle =\int_1^3(x^3-4x)dx+\int_3^2(x^3-4x)dx\)
\(\displaystyle =\int_{1}^2(x^3-4x)dx\)
\(\displaystyle =\left[\frac{1}{4}x^4-2x^2\right]_1^2\)
\(\displaystyle =-4+\frac{7}{4}\)
\(\displaystyle =-\frac{9}{4}\)
2.次の等式をみたす関数\(f(x)\)を求めなさい。
(1)\(\displaystyle f(x)=3x^2+\int_{-1}^1f(t)dt\)
\(\displaystyle a=\int_{-1}^1f(t)dt\)とおく。
\(f(x)=3x^2+a\)
\(f(t)=3t^2+a\)
\(\displaystyle a=\int_{-1}^1f(t)dt\)
\(\displaystyle \ \ =\int_{-1}^1(3t^2+a)dt\)
\(\ \ =[t^3+at]_{-1}^1\)
\(\ \ =2a+2\)
\(a=-2\)
よって、
\(f(x)=3x^2-2\)
(2)\(\displaystyle f(x)=x^2+4x-\int_0^1f(t)dt\)
\(\displaystyle a=\int_0^1f(t)dt\)とおく。
\(f(x)=x^2+4x-a\)
\(f(t)=t^2+4t-a\)
\(\displaystyle a=\int_0^1f(t)dt\)
\(\displaystyle \ \ =\int_0^1(t^2+4t-a)dt\)
\(\displaystyle \ \ =\left[\frac{1}{3}t^3+2t^2-at\right]_0^1\)
\(\displaystyle \ \ =-a+\frac{7}{3}\)
\(\displaystyle a=\frac{7}{6}\)
よって、
\(\displaystyle f(x)=x^2+4x-\frac{7}{6}\)
(3)\(\displaystyle f(x)=2x^2+2x-\int_{-3}^0f(t)dt\)
\(\displaystyle a=\int_{-3}^0f(t)dt\)とおく。
\(f(x)=2x^2+2x-a\)
\(f(t)=2t^2+2t-a\)
\(\displaystyle a=\int_{-3}^0f(t)dt\)
\(\displaystyle \ \ =\int_{-3}^0(2t^2+2t-a)dt\)
\(\displaystyle \ \ =\left[\frac{2}{3}t^3+t^2-at\right]_{-3}^0\)
\(\ \ =-3a+9\)
\(\displaystyle a=\frac{9}{4}\)
よって、
\(\displaystyle f(x)=2x^2+2x-\frac{9}{4}\)
3.次の等式をみたす関数\(f(x)\)と定数\(a\)を求めなさい。
(1)\(\displaystyle \int_a^x f(t)dt=x^2-x-2\)
両辺を微分すると、
\(f(x)=2x-1\)
また、\(x=a\)とおくと、左辺は\(0\)になるので、
\(a^2-a-2=0\)
\((a+1)(a-2)=0\)
\(a=-1,2\)
(2)\(\displaystyle \int_a^x f(t)dt=x^2-4x-12\)
両辺を微分すると、
\(f(x)=2x-4\)
また、\(x=a\)とおくと、左辺は\(0\)になるので、
\(a^2-4a-12=0\)
\((a+2)(a-6)=0\)
\(a=-2,6\)
(3)\(\displaystyle \int_a^x f(t)dt=3x^2+4x+1\)
両辺を微分すると、
\(f(x)=6x+4\)
また、\(x=a\)とおくと、左辺は\(0\)になるので、
\(6a^2+4a+1=0\)
\((3a+1)(a+1)=0\)
\(\displaystyle a=-\frac{1}{3},-1\)
