2022학년도 11월 미적분 30번

30. 실수 전체의 집합에서 증가하고 미분가능한 함수 $f(x)$$f(x)$가 다음 조건을 만족시킨다.

(가) $f(1)=1,\begin{array}{r}{\int }_1^{2}f(x)dx=\frac{5}{4}\end{array}$$f(1)=1,~\begin{aligned} \int_1^2 f(x)dx=\cfrac 54\end{aligned}$

(나) 함수 $f(x)$$f(x)$의 역함수를 $g(x)$$g(x)$라 할 때, $x\ge 1$$x\ge 1$인 모든 실수 $x$$x$에 대하여 $g(2x)=2f(x)$$g(2x)=2f(x)$이다.

$\begin{array}{r}{\int }_1^{8}x{f}^{\prime }(x)dx=\frac{q}{p}\end{array}$$\begin{aligned} \int_1^8 xf'(x)dx =\cfrac qp\end{aligned}$일 때, $p+q$$p+q$의 값을 구하시오. (단, $p$$p$와 $q$$q$는 서로소인 자연수이다.)

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i. 정리

• $f^{\prime }(x)\ge 0$$f'(x)\ge 0$

• $f(1)=1,\begin{array}{r}{\int }_1^{2}f(x)dx=\frac{5}{4}\end{array}$$f(1)=1,~\begin{aligned} \int_1^2 f(x)dx=\cfrac 54\end{aligned}$

• $g(x)=f^{-1}(x),x\ge 1⟶g(2x)=2f(x)$$g(x)=f^{-1}(x),~x\ge 1\quad\longrightarrow\quad g(2x)=2f(x)$

  어랏? 뭐지?

• $\begin{array}{r}{\int }_1^{8}x{f}^{\prime }(x)dx=[xf(x){]}_1^8-{\int }_1^{8}f(x)dx\end{array}$$\begin{aligned}\int_1^8 xf'(x)dx =\Big[xf(x)\Big]_1^8 -\int_1^8 f(x)dx \end{aligned}$

ii. 생각

• $f(g(2x))=f(2f(x))=2x$$f\big(g(2x)\big)=f\big(2f(x)\big)=2x$

  • $x=1$$x=1$이면, $f(2)=2$$f(2)=2$
  • $x=2$$x=2$이면, $f(4)=4$$f(4)=4$
  • $x=3$$x=3$이면, $f(2f(3))=6$$f(2f(3))=6$ 이건 아닌 듯?
  • $x=4$$x=4$이면, $f(8)=8$$f(8)=8$

  뭘까?

  우선 $[xf(x){]}_1^8=8f(8)-f(1)=63$$\Big[xf(x)\Big]_1^8=8f(8)-f(1)=63$

• $\begin{array}{rl}{\int }_1^{8}f(x)dx& ={\int }_1^{2}f(x)dx+{\int }_2^{4}f(x)dx+{\int }_4^{8}f(x)dx\end{array}$$\begin{aligned}\int_1^8f(x)dx &=\int_1^2f(x)dx+\int_2^4f(x)dx+\int_4^8f(x)dx \end{aligned}$로 구하라는 건가?

iii. 또 생각하자.

• $\begin{array}{r}{\int }_2^{4}f(x)dx=?\end{array}$$\begin{aligned} \int_2^4f(x)dx=?\end{aligned}$

  $g(2x)=2f(x)$$g(2x)=2f(x)$를 이용해봐야겠지?

  양변을 적분해보자.

  $\begin{array}{r}{\int }_2^{4}g(2x)dx=2{\int }_2^{4}f(x)dx\end{array}$$\begin{aligned} \int_2^4g(2x)dx=2\int_2^4f(x)dx\end{aligned}$

  $2x=t$$2x=t$ 로 치환을 해보자.

  $2dx=dt,4\to 8$$2dx=dt, 4\rightarrow 8$

  $\begin{array}{r}\frac{1}{2}{\int }_4^{8}g(t)dt=2{\int }_2^{4}f(x)dx\end{array}$$\begin{aligned}\cfrac 12\int_4^8 g(t)dt=2\int_2^4 f(x)dx \end{aligned}$

  어? 어랏?

  $\begin{array}{r}{\int }_2^{4}f(x)dx=\frac{1}{4}{\int }_4^{8}g(t)dt\end{array}$$\begin{aligned} \int_2^4 f(x)dx =\cfrac 14 \int_4^8 g(t)dt\end{aligned}$

  $g(x)$$g(x)$는 $f(x)$$f(x)$의 역함수....

  $\begin{array}{rl}{\int }_4^{8}g(t)dt& =8^2-4^2-{\int }_4^{8}f(x)dx\\ \\ & =48-{\int }_4^{8}f(x)dx\end{array}$$\begin{aligned}\int_4^8 g(t)dt&=8^2 -4^2 -\int_4^8f(x)dx \\ \\ &= 48-\int_4^8 f(x)dx\end{aligned}$

  $\begin{array}{rl}{\int }_2^{4}f(x)dx& =12-\frac{1}{4}{\int }_4^{8}f(x)dx\\ \\ {\int }_4^{8}f(x)dx& =48-4{\int }_2^{4}f(x)dx\end{array}$$\begin{aligned}\int_2^4 f(x)dx &=12-\cfrac 14\int_4^8 f(x)dx \\ \\ \int_4^8 f(x)dx &=48-4\int_2^4 f(x)dx\end{aligned}$

어랏? $2$$2$부터 $4$$4$까지 적분을 하니까 다음 과정이 얻어졌다? 그럼 풀렸네.

• $\begin{array}{r}{\int }_1^{2}f(x)dx=\frac{5}{4}\end{array}$$\begin{aligned} \int_1^2f(x)dx =\cfrac 54\end{aligned}$를 이용하자.

  $\begin{array}{rl}{\int }_1^{2}g(x)dx& =2{\int }_1^{2}f(x)dx\\ \\ \frac{1}{2}{\int }_2^{4}g(t)dt& =\frac{5}{2}\\ \\ {\int }_2^{4}g(t)dt& =5\\ \\ 4^2-2^2-{\int }_2^{4}f(x)dx& =5\end{array}$$\begin{aligned}\int_1^2 g(x)dx& =2\int_1^2 f(x)dx \\ \\ \cfrac 12 \int_2^4g(t)dt &=\cfrac 52 \\ \\ \int_2^4g(t)dt&=5 \\ \\ 4^2 -2^2 -\int_2^4 f(x)dx &=5 \end{aligned}$

  $\begin{array}{rl}{\int }_2^{4}f(x)dx& =12-5\\ \\ & =7\end{array}$$\begin{aligned} \int_2^4 f(x)dx &=12-5\\ \\ &=7\end{aligned}$

정리하면,

$\begin{array}{rl}{\int }_1^{2}f(x)dx& =\frac{5}{4}\\ \\ {\int }_2^{4}f(x)dx& =7\\ \\ {\int }_4^{8}f(x)dx& =20\end{array}$$\begin{aligned} \int_1^2 f(x)dx &=\cfrac 54\\ \\ \int_2^4 f(x)dx &=7\\ \\ \int_4^8 f(x)dx &=20 \end{aligned}$

iv. 계산하자.

$\begin{array}{rl}{\int }_1^{8}x{f}^{\prime }(x)dx& =63-(\frac{5}{4}+7+20)\\ \\ & =\frac{139}{4}\end{array}$$\begin{aligned} \int_1^8 xf'(x)dx &= 63-(\cfrac 54 +7+20)\\ \\ &=\cfrac {139}4\end{aligned}$

$\therefore p+q=143$$\therefore~p+q=143$