北海道大学 1988年度 前期日程 第2次学力試験文系数学 第4問

(1)
$0\leqq x\leqq 2$ では$$\begin{array}{rl}{\displaystyle f(x)}& {\displaystyle ={\int }_0^x{t}^2(x-t)dt+{\int }_x^2{t}^2(t-x)dt}\\ {\displaystyle }& {\displaystyle ={\displaystyle \frac{x^4}{12}}+(4-{\displaystyle \frac{8x}{3}}+{\displaystyle \frac{x^4}{12}})}\\ {\displaystyle }& {\displaystyle ={\displaystyle \frac{x^4}{6}}-{\displaystyle \frac{8x}{3}}+4.}\end{array}$$$x\geqq 2$ では全ての $0\leqq t\leqq 2$ に対して $t-x\leqq 0$ だから$$\begin{array}{rl}{\displaystyle f(x)}& {\displaystyle ={\int }_0^2{t}^2(x-t)dt}\\ {\displaystyle }& {\displaystyle ={\displaystyle \frac{8x}{3}}-4.}\end{array}$$したがって$$f(x)=\{\begin{array}{ll}{\displaystyle \frac{x^4}{6}}-{\displaystyle \frac{8x}{3}}+4& (0\leqq x\leqq 2),\\ {\displaystyle \frac{8x}{3}}-4& (x\geqq 2)\mathrm{.}\end{array}$$両式は $x=2$ でともに $4\mathrm{/}3$ となる。

(2) $$\begin{array}{rl}{\displaystyle {\int }_0^{3}f(x)dx=}& {\displaystyle {\int }_0^2({\displaystyle \frac{x^4}{6}}-{\displaystyle \frac{8x}{3}}+4)dx}\\ {\displaystyle }& {\displaystyle +{\int }_2^3({\displaystyle \frac{8x}{3}}-4)dx}\\ {\displaystyle =}& {\displaystyle {\displaystyle \frac{56}{15}}+{\displaystyle \frac{8}{3}}}\\ {\displaystyle =}& {\displaystyle {\displaystyle \frac{32}{5}}\mathrm{.}}\end{array}$$

(1) は絶対値を含む積分を $x$ で微分し、初期値から $f$ を復元する。(2) は二重積分として $x,t$ の順序を交換し、内側の ${\int }_0^3\mathrm{\mid }t-x\mathrm{\mid }dx$ を先に求める。

(1)
$0<x<2$ では$$\begin{array}{rl}{\displaystyle f^{\mathrm{\prime }}(x)}& {\displaystyle ={\int }_0^x{t}^{2}dt-{\int }_x^2{t}^{2}dt}\\ {\displaystyle }& {\displaystyle ={\displaystyle \frac{2x^3}{3}}-{\displaystyle \frac{8}{3}}\mathrm{.}}\end{array}$$また$$f(0)={\int }_0^2{t}^{3}dt=4.$$したがって$$f(x)=4+{\int }_0^x({\displaystyle \frac{2u^3}{3}}-{\displaystyle \frac{8}{3}})du={\displaystyle \frac{x^4}{6}}-{\displaystyle \frac{8x}{3}}+4.$$$x>2$ では常に $x-t>0$ なので$$f^{\mathrm{\prime }}(x)={\int }_0^2{t}^{2}dt={\displaystyle \frac{8}{3}}\mathrm{.}$$$f(2)=4\mathrm{/}3$ を用いれば$$f(x)={\displaystyle \frac{4}{3}}+{\displaystyle \frac{8}{3}}(x-2)={\displaystyle \frac{8x}{3}}-4.$$(2)
積分領域は長方形$$0\leqq x\leqq 3,0\leqq t\leqq 2$$なので、積分の順序を交換して$$\begin{array}{rl}{\displaystyle {\int }_0^{3}f(x)dx}& {\displaystyle ={\int }_0^2{t}^2\{{\int }_0^3\mathrm{\mid }t-x\mathrm{\mid }dx\}dt\mathrm{.}}\end{array}$$$0\leqq t\leqq 2$ では$$\begin{array}{rl}{\displaystyle {\int }_0^3\mathrm{\mid }t-x\mathrm{\mid }dx}& {\displaystyle ={\int }_0^t(t-x)dx+{\int }_t^3(x-t)dx}\\ {\displaystyle }& {\displaystyle ={\displaystyle \frac{t^2}{2}}+{\displaystyle \frac{(3-t{)}^2}{2}}\mathrm{.}}\end{array}$$よって$$\begin{array}{rl}{\displaystyle {\int }_0^{3}f(x)dx}& {\displaystyle ={\displaystyle \frac{1}{2}}{\int }_0^2{t}^2\{t^2+(3-t{)}^2\}dt}\\ {\displaystyle }& {\displaystyle ={[{\displaystyle \frac{t^5}{5}}-{\displaystyle \frac{3t^4}{4}}+{\displaystyle \frac{3t^3}{2}}]}_0^2}\\ {\displaystyle }& {\displaystyle ={\displaystyle \frac{32}{5}}\mathrm{.}}\end{array}$$