2019학년도 06월 가형 30번

30. 실수 전체의 집합에서 미분가능한 함수 $f(x)$$f(x)$에 대하여 곡선 $y=f(x)$$y=f(x)$ 위의 점 $(t,f(t))$$(t,~f(t))$에서의 접선의 $y$$y$ 절편을 $g(t)$$g(t)$라 하자. 모든 실수 $t$$t$에 대하여

$$(1+t^2)\{g(t+1)-g(t)\}=2t$$

이고, $\begin{array}{r}{\int }_0^{1}f(x)dx=-\frac{\ln 10}{4},f(1)=4+\frac{\ln 17}{8}\end{array}$$\begin{aligned} \int_0^1 f(x)dx =-\cfrac{\ln 10}4,~f(1)=4+\cfrac{\ln 17}8\end{aligned}$일 때, $\begin{array}{r}2\{f(4)+f(-4)\}-{\int }_{-4}^{4}f(x)dx\end{array}$$\begin{aligned} 2\big\{f(4)+f(-4)\big\} -\int_{-4}^4f(x)dx \end{aligned}$의 값을 구하시오.

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i. 정리

• $y=f(x)$$y=f(x)$는 미분가능
• $(t,f(t))$$(t,~f(t))$에서의 접선의 $y$$y$ 절편 : $g(t)$$g(t)$
• $(1+t^2)\{g(t+1)-g(t)\}=2t$$(1+t^2)\big\{g(t+1)-g(t)\big\} =2t$
• $\begin{array}{r}{\int }_0^{1}f(x)dx=-\frac{\ln 10}{4},f(1)=4+\frac{\ln 17}{8}\end{array}$$\begin{aligned} \int_0^1 f(x)dx =-\cfrac{\ln 10}4,~f(1)=4+\cfrac{\ln 17}8\end{aligned}$

ii. 생각

• 우선 가능한 $g(t)$$g(t)$를 구하자.

  $y=f^{\prime }(t)(x-t)+f(t)$$y=f'(t)(x-t)+f(t)$

  $g(t)=f(t)-t{f}^{\prime }(t)$$g(t)=f(t)-tf'(t)$

• $g(t)$$g(t)$를 $0$$0$부터 $1$$1$까지 적분하면, $\begin{array}{r}{\int }_0^{1}f(x)dx\end{array}$$\begin{aligned} \int_0^1 f(x)dx\end{aligned}$를 이용할 수 있을 듯 하다?

  $\begin{array}{rl}{\int }_0^{1}g(t)dt& ={\int }_0^{1}f(t)dt-[tf(t){]}_0^1+{\int }_0^{1}f(t)dt\\ \\ & =2{\int }_0^{1}f(t)dt-[tf(t){]}_0^1\\ \\ & =2{\int }_0^{1}f(t)dt-f(1)\\ \\ & =-\frac{\ln 10}{2}-4-\frac{\ln 17}{8}\end{array}$$\begin{aligned} \int_0^1 g(t)dt &= \int_0^1 f(t)dt -\big[tf(t)\big]_0^1 +\int_0^1 f(t)dt \\ \\ &= 2\int_0^1 f(t)dt -\big[tf(t)\big]_0^1 \\ \\ &=2\int_0^1 f(t)dt -f(1)\\ \\ &= -\cfrac {\ln 10}2-4-\cfrac {\ln 17}8\end{aligned}$

• 어? $g(t)$$g(t)$을 $-4$$-4$부터 $4$$4$까지 적분하면 $\begin{array}{r}{\int }_{-4}^{4}f(t)dt\end{array}$$\begin{aligned} \int_{-4}^4f(t)dt\end{aligned}$의 꼴은 꺼낼 수 있을 듯 하다.

  $\begin{array}{rl}{\int }_{-4}^{4}g(t)dt& =2{\int }_{-4}^{4}f(t)dt-[tf(t){]}_{-4}^4\\ \\ & =2{\int }_{-4}^{4}f(t)dt-(4f(4)+4f(-4))\\ \\ & =2{\int }_{-4}^{4}f(t)dt-4\{f(4)+f(-4)\}\end{array}$$\begin{aligned}\int_{-4}^4 g(t)dt &= 2\int_{-4}^4 f(t)dt -\Big[tf(t)\Big]_{-4}^4 \\ \\ &= 2\int_{-4}^4f(t)dt -\big(4f(4)+4f(-4)\big)\\ \\ &= 2\int_{-4}^4 f(t)dt-4\big\{f(4)+f(-4)\big\} \end{aligned}$

  어랏? 다 튀어나오네?

  $\begin{array}{r}2\{f(4)+f(-4)\}-{\int }_{-4}^{4}f(t)dt=-\frac{1}{2}{\int }_{-4}^{4}g(t)dt\end{array}$$\begin{aligned} 2\big\{f(4)+f(-4)\big\}-\int_{-4}^4 f(t)dt=-\cfrac 12 \int_{-4}^4 g(t)dt\end{aligned}$

• $\begin{array}{r}g(t+1)-g(t)=\frac{2t}{1+t^2}\end{array}$$\begin{aligned} g(t+1)-g(t)=\cfrac {2t}{1+t^2}\end{aligned}$를 이용하면 될 거 같은데?

  $\begin{array}{rl}{\int }_n^{n+1}g(t+1)dt-{\int }_n^{n+1}g(t)& ={\int }_n^{n+1}\frac{2t}{1+t^2}dt\end{array}$$\begin{aligned} \int_n^{n+1} g(t+1)dt -\int_n^{n+1} g(t)&=\int_n^{n+1} \cfrac{2t}{1+t^2}dt \end{aligned}$

  식을 좀 변형하자.

  $\begin{array}{rl}{\int }_n^{n+1}g(t+1)dt& ={\int }_n^{n+1}\frac{2t}{1+t^2}dt+{\int }_n^{n+1}g(t)dt\end{array}$$\begin{aligned} \int_n^{n+1}g(t+1)dt &=\int_n^{n+1}\cfrac {2t}{1+t^2}dt +\int_n^{n+1}g(t)dt \end{aligned}$

  • $\begin{array}{r}{\int }_n^{n+1}g(t+1)dt\end{array}$$\begin{aligned} \int_n^{n+1} g(t+1)dt\end{aligned}$를 치환하자.

    $t+1=u⟶dt=du,n+1\to n+2$$t+1=u\quad \longrightarrow \quad dt=du,~n+1\to n+2$

    $\begin{array}{r}{\int }_n^{n+1}g(t+1)dt={\int }_{n+1}^{n+2}g(u)du\end{array}$$\begin{aligned} \int_n^{n+1} g(t+1)dt =\int_{n+1}^{n+2}g(u)du \end{aligned}$

  • $\begin{array}{r}{\int }_n^{n+1}\frac{2t}{1+t^2}dt\end{array}$$\begin{aligned} \int_n^{n+1}\cfrac {2t}{1+t^2}dt\end{aligned}$를 적분해놓자.

    $1+t^2=k⟶2tdt=dk,1+n^2\to 1+(n+1{)}^2$$1+t^2 =k\quad \longrightarrow \quad 2tdt=dk,~1+n^2 \to 1+(n+1)^2$

    $\begin{array}{rl}\begin{array}{c}{\int }_n^{n+1}\frac{2t}{1+t^2}dt\end{array}& ={\int }_{1+n^2}^{1+(n+1{)}^2}\frac{1}{k}dk\\ \\ & =\ln \frac{1+(1+n{)}^2}{1+n^2}\end{array}$$\begin{aligned} \begin{aligned} \int_n^{n+1}\cfrac {2t}{1+t^2}dt\end{aligned} &= \int_{1+n^2}^{1+(n+1)^2} \cfrac 1k dk \\ \\ &= \ln \cfrac {1+(1+n)^2}{1+n^2} \end{aligned}$

• $\begin{array}{rl}{\int }_n^{n+1}g(t+1)dt& ={\int }_n^{n+1}\frac{2t}{1+t^2}dt+{\int }_n^{n+1}g(t)dt\end{array}$$\begin{aligned} \int_n^{n+1}g(t+1)dt &=\int_n^{n+1}\cfrac {2t}{1+t^2}dt +\int_n^{n+1}g(t)dt \end{aligned}$를 다시 쓰면,

  $\begin{array}{rl}{\int }_{n+1}^{n+2}g(t)dt& ={\int }_n^{n+1}\frac{2t}{1+t^2}dt+{\int }_n^{n+1}g(t)dt\\ \\ & ={\int }_n^{n+1}g(t)dt+\ln \frac{1+(1+n{)}^2}{1+n^2}\end{array}$$\begin{aligned} \int_{n+1}^{n+2}g(t)dt &=\int_n^{n+1}\cfrac {2t}{1+t^2}dt +\int_n^{n+1}g(t)dt \\ \\ &=\int_n^{n+1}g(t)dt +\ln \cfrac {1+(1+n)^2}{1+n^2}\end{aligned}$

  와...이제부터 산수가 시작되겠어...

   

  • $n=0$$n=0$이면

    $\begin{array}{rl}{\int }_1^{2}g(t)dt& ={\int }_0^{1}g(t)dt+\ln 2\end{array}$$\begin{aligned} \int_1^2 g(t)dt &=\int_0^1 g(t)dt +\ln 2\end{aligned}$

  • $n=1$$n=1$이면

    $\begin{array}{rl}{\int }_2^{3}g(t)dt& ={\int }_1^{2}g(t)dt+\ln \frac{5}{2}\\ \\ & ={\int }_0^{1}g(t)dt+\ln 5\end{array}$$\begin{aligned} \int_2^3 g(t)dt &=\int_1^2 g(t)dt +\ln \cfrac 52 \\ \\ &= \int_0^1 g(t)dt +\ln 5\end{aligned}$

  • $n=2$$n=2$이면

    $\begin{array}{rl}{\int }_3^{4}g(t)dt& ={\int }_2^{3}g(t)dt+\ln \frac{10}{5}\\ \\ & ={\int }_0^{1}g(t)dt+\ln 10\end{array}$$\begin{aligned} \int_3^4 g(t)dt &=\int_2^3 g(t)dt +\ln\cfrac {10}5\\ \\ &= \int_0^1 g(t)dt +\ln 10 \end{aligned}$

  왠지 규칙이 있는 거 같다? 아무튼 구한 걸 정리하면,

  $\begin{array}{r}{\int }_0^{4}g(t)dt=4{\int }_0^{1}g(t)dt+2\ln 10\end{array}$$\begin{aligned} \int_0^4 g(t)dt =4\int_0^1 g(t)dt +2\ln 10\end{aligned}$

   

  식을 조금 바꾸자.

  $\begin{array}{r}{\int }_n^{n+1}g(t)dt={\int }_{n+1}^{n+2}g(t)dt-\ln \frac{1+(1+n{)}^2}{1+n^2}\end{array}$$\begin{aligned}\int_n^{n+1} g(t)dt =\int_{n+1}^{n+2}g(t)dt -\ln \cfrac{1+(1+n)^2}{1+n^2}\end{aligned}$

   

  • $n=-1$$n=-1$을 대입하면,

    $\begin{array}{rl}{\int }_{-1}^{0}g(t)dt& ={\int }_0^{1}g(t)dt-\ln \frac{1}{2}\\ \\ & ={\int }_0^{1}g(t)dt+\ln 2\end{array}$$\begin{aligned} \int_{-1}^{0} g(t)dt &=\int_{0}^{1}g(t)dt -\ln \cfrac{1}{2}\\ \\ &=\int_0^1g(t)dt +\ln 2 \end{aligned}$

  • $n=-2$$n=-2$를 대입하면,

    $\begin{array}{rl}{\int }_{-2}^{-1}g(t)dt& ={\int }_{-1}^{0}g(t)dt-\ln \frac{2}{5}\\ \\ & ={\int }_{-1}^{0}g(t)dt+\ln \frac{5}{2}\\ \\ & ={\int }_0^{1}g(t)dt+\ln 5\end{array}$$\begin{aligned} \int_{-2}^{-1} g(t)dt &=\int_{-1}^{0}g(t)dt -\ln \cfrac{2}{5}\\ \\ &=\int_{-1}^{0}g(t)dt +\ln \cfrac{5}{2} \\ \\ &=\int_0^1 g(t)dt+\ln 5 \end{aligned}$

  • $n=-3$$n=-3$을 대입하면,

    규칙있네...

    $\begin{array}{r}{\int }_{-3}^{-2}g(t)dt={\int }_0^{1}g(t)dt+\ln 10\end{array}$$\begin{aligned} \int_{-3}^{-2} g(t)dt =\int_0^1 g(t)dt +\ln 10\end{aligned}$

  • $n=-4$$n=-4$를 대입하면,

    $\begin{array}{r}{\int }_{-4}^{-3}g(t)dt={\int }_0^{1}g(t)dt+\ln 17\end{array}$$\begin{aligned} \int_{-4}^{-3}g(t)dt =\int_0^1 g(t)dt +\ln 17\end{aligned}$

  $\begin{array}{r}{\int }_{-4}^{0}g(t)dt=4{\int }_0^{1}g(t)dt+2\ln 10+\ln 17\end{array}$$\begin{aligned} \int_{-4}^0g(t)dt =4\int_0^1 g(t)dt +2\ln 10+\ln 17\end{aligned}$

• 마지막 계산을 하자.

  $\begin{array}{rl}{\int }_{-4}^{4}g(t)dt& ={\int }_{-4}^{0}g(t)dt+{\int }_0^{4}g(t)dt\\ \\ & =8{\int }_0^{1}g(t)dt+4\ln 10+\ln 17\\ \\ & =-4\ln 10-32-\ln 17+4\ln 10+\ln 17\\ \\ & =-32\end{array}$$\begin{aligned} \int_{-4}^4 g(t)dt &=\int_{-4}^0g(t)dt+\int_0^4 g(t)dt \\ \\ &= 8\int_0^1 g(t)dt +4\ln 10 +\ln 17 \\ \\ &= -4\ln 10-32-\ln 17+4\ln 10 +\ln 17 \\ \\ &=-32\end{aligned}$

$\therefore \begin{array}{r}2\{f(4)+f(-4)\}-{\int }_{-4}^{4}f(t)dt=-\frac{1}{2}{\int }_{-4}^{4}g(t)dt=16\end{array}$$\therefore~ \begin{aligned} 2\big\{f(4)+f(-4)\big\}-\int_{-4}^4 f(t)dt=-\cfrac 12 \int_{-4}^4 g(t)dt=16\end{aligned}$