2019년 10월 가형 30번

30. 실수 전체의 집합에서 미분가능한 두 함수 $f(x),g(x)$$f(x),~g(x)$가 모든 실수 $x$$x$에 대하여 다음 조건을 만족시킨다.

(가) $g(x+1)-g(x)=-\pi (e+1)e^x\sin (\pi x)$$g(x+1)-g(x)=-\pi (e+1)e^x \sin (\pi x)$

(나) $\begin{array}{rl}g(x+1)& ={\int }_0^x\{f(t+1)e^t-f(t)e^t+g(t)\}dt\end{array}$$\begin{aligned} g(x+1)&=\int_0^x \Big\{f(t+1)e^t -f(t)e^t +g(t)\Big\} dt\end{aligned}$

$\begin{array}{r}{\int }_0^{1}f(x)dx=\frac{10}{9}e+4\end{array}$$\begin{aligned} \int_0^1 f(x)dx =\cfrac {10}9 e +4\end{aligned}$일 때, $\begin{array}{r}{\int }_0^{1}f(x)dx\end{array}$$\begin{aligned} \int_0^1 f(x)dx \end{aligned}$의 값을 구하시오.

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i. 생각

• 우선 습관적으로 $x=0$$x=0$을 이용하면,

  $g(1)=0$$g(1)=0$ ( 조건 (나))

  $g(1)-g(0)=0⟶g(0)=g(1)=0$$g(1)-g(0)=0\quad \longrightarrow \quad g(0)=g(1)=0$

  어? $\sin (\pi x)$$\sin (\pi x)$는 모든 정수에 대해서 $0$$0$이네?

  $g(n)=0$$g(n)=0$ (단, $n$$n$은 정수)

• 아무래도 식을 미분하고 적분하고 장난쳐서 어떻게든 $g(x)$$g(x)$를 소거시켜야할 거 같다.

  • (가) 를 미분하면,

    $g^{\prime }(x+1)-g^{\prime }(x)=-\pi (e+1)e^x\sin \pi x-{\pi }^2(e+1)e^x\cos \pi x$$g'(x+1)-g'(x)=-\pi (e+1)e^x \sin \pi x -\pi^2 (e+1)e^x \cos \pi x$

    깔끔하게 정리하면,

    $g^{\prime }(x+1)-g^{\prime }(x)=-\pi (e+1)e^x(\sin \pi x+\pi \cos \pi x)$$g'(x+1)-g'(x)=-\pi(e+1)e^x (\sin \pi x +\pi \cos \pi x)$

  • (나)를 미분하면,

    $g^{\prime }(x+1)=f(x+1)e^x-f(x)e^x+g(x)$$g'(x+1)=f(x+1)e^x -f(x)e^x +g(x)$

    이 식을 정리하면 $f(x+1)-f(x)$$f(x+1)-f(x)$을 구할 수 있겠다.

    $f(x+1)-f(x)=e^{-x}\{g^{\prime }(x+1)-g(x)\}$$f(x+1)-f(x)=e^{-x} \big\{ g'(x+1)-g(x)\big\}$

• 에이..설마...에이...아닐꺼야...

  $g^{\prime }(x+1)=g^{\prime }(x)-\pi (e+1)e^x(\sin \pi x+\pi \cos \pi x)$$g'(x+1)=g'(x)-\pi (e+1)e^x (\sin \pi x +\pi \cos\pi x)$를 대입하자...

  $\begin{array}{rl}f(x+1)-f(x)& =e^{-x}\{g^{\prime }(x)-g(x)-\pi (e+1)e^x(\sin \pi x+\pi \cos \pi x)\\ \\ & =e^{-x}\{g^{\prime }(x)-g(x)\}-\pi (e+1)(\sin \pi x+\pi \cos \pi x)\end{array}$$\begin{aligned}f(x+1)-f(x)&= e^{-x} \big\{ g'(x)-g(x)-\pi (e+1)e^x (\sin \pi x +\pi \cos \pi x ) \\ \\ &= e^{-x} \big\{ g'(x)-g(x)\big\} -\pi (e+1)(\sin\pi x +\pi \cos \pi x) \end{aligned}$

  어디보자....

  $\{g(x)e^{-x}{\}}^{\prime }=g^{\prime }(x)e^{-x}-g(x)e^{-x}$$\Big\{g(x)e^{-x}\Big\}'=g'(x)e^{-x} -g(x)e^{-x}$

  아..적분하라는 소리군...

• 적분하자.

  $\begin{array}{rl}{\int }_x^{x+1}\{f(x+1)-f(x)\}dx& =[g(x)e^{-x}{]}_x^{x+1}+(e+1)[\cos \pi x{]}_x^{x+1}-\pi (e+1)[\sin \pi x{]}_x^{x+1}\end{array}$$\begin{aligned} \int_x^{x+1} \big\{f(x+1)-f(x)\big\}dx &= \Big[g(x)e^{-x} \Big]_{x}^{x+1} +(e+1)\Big[\cos \pi x \Big]_x^{x+1}-\pi (e+1)\Big[\sin\pi x \Big]_x^{x+1} \end{aligned}$

  $\begin{array}{rl}{\int }_x^{x+1}f(x+1)dx-{\int }_x^{x+1}f(x)dx& =\sim \end{array}$$\begin{aligned} \int_x^{x+1} f(x+1)dx -\int_x^{x+1} f(x)dx &=\sim \end{aligned}$

  그리고 $[\sin \pi x{]}_x^{x+1}$$\Big[\sin\pi x \Big]_x^{x+1}$은 $x$$x$가 정수이면 $0$$0$이다!!!!

   

  $\begin{array}{rl}{\int }_x^{x+1}f(x)dx& ={\int }_{x-1}^{x}f(x)dx-(e+1)[\cos \pi x{]}_x^{x+1}\end{array}$$\begin{aligned} \int_x^{x+1} f(x)dx &=\int_{x-1}^x f(x)dx -(e+1)\Big[\cos \pi x \Big]_x^{x+1} \end{aligned}$

• $x=1$$x=1$부터 대입해나가면 될 거 같은데?

  • $x=1$$x=1$

    $\begin{array}{rl}{\int }_1^{2}f(x)dx& ={\int }_0^{1}f(x)dx-2(e+1)\end{array}$$\begin{aligned}\int_1^2 f(x)dx &= \int_0^1 f(x)dx -2(e+1) \end{aligned}$

  • $x=2$$x=2$

    $\begin{array}{rl}{\int }_2^{3}f(x)dx& ={\int }_1^{2}f(x)dx+2(e+1)\\ \\ & ={\int }_0^{1}f(x)dx\end{array}$$\begin{aligned}\int_2^3 f(x)dx &= \int_1^2 f(x)dx +2(e+1)\\ \\ &=\int_0^1f(x)dx \end{aligned}$

    오...반복인가?

  • $x=3$$x=3$

    $\begin{array}{rl}{\int }_3^{4}f(x)dx& ={\int }_2^{3}f(x)dx-2(e+1)\\ \\ & ={\int }_0^{1}f(x)dx-2(e+1)\end{array}$$\begin{aligned}\int_3^4 f(x)dx &= \int_2^3 f(x)dx -2(e+1) \\ \\ &= \int_0^1 f(x)dx -2(e+1) \end{aligned}$

  오호라 반복이다!

• 계산하자.

  $\begin{array}{rl}{\int }_1^{10}f(x)dx& =9{\int }_0^{1}f(x)dx-5\times 2(e+1)\\ \\ & =10e+36-10e-10\\ \\ & =26\end{array}$$\begin{aligned}\int_1^{10} f(x)dx &= 9\int_0^1 f(x)dx -5\times 2(e+1)\\ \\ &= 10e+36 -10e-10\\ \\ &= 26 \end{aligned}$

$\therefore 26$$\therefore~ 26$