주기 우함수 부분적분

[출제의도] 부분적분법 이해하기 $\displaystyle \int _{- 1} ^{5} {} f \left( x \right) \left( x + \cos 2 \pi x \right) dx$ $\displaystyle = \int _{- 1} ^{5} {} xf \left( x \right) dx + \int _{- 1} ^{5} {} f \left( x \right) \cos 2 \pi x dx$ $\cdots$ ㉠ 조건 (가)에 의하여 $\displaystyle \int _{- 1} ^{1} {xf \left( x \right) dx} = 0$, $\displaystyle \int _{- 1} ^{1} {f \left( x \right) dx = 2 \int _{0} ^{1} {f \left( x \right) dx}}$ $\displaystyle \int _{- 1} ^{5} {} xf \left( x \right) dx$ $\displaystyle = \int _{- 1} ^{1} {} xf \left( x \right) dx + \int _{1} ^{3} {} xf \left( x \right) dx + \int _{3} ^{5} {} xf \left( x \right) dx$ $\displaystyle = \int _{- 1} ^{1} {} xf \left( x \right) dx + \int _{- 1} ^{1} {} \left( x + 2 \right) f \left( x + 2 \right) dx$$\displaystyle + \int _{- 1} ^{1} {} \left( x + 4 \right) f \left( x + 4 \right) dx$ $\displaystyle = \int _{- 1} ^{1} {} xf \left( x \right) dx + \int _{- 1} ^{1} {} \left( x + 2 \right) f \left( x \right) dx$$\displaystyle + \int _{- 1} ^{1} {} \left( x + 4 \right) f \left( x \right) dx$ $\displaystyle = 3 \int _{- 1} ^{1} {} xf \left( x \right) dx + 6 \int _{- 1} ^{1} {} f \left( x \right) dx$ $\displaystyle = 12 \int _{0} ^{1} {} f \left( x \right) dx = 24$ $\cdots$ ㉡ 조건 (가), (나)에 의하여 모든 실수 $x$에 대하여 $f \left( - x \right) \cos 2 \pi \left( - x \right) \mathit{=} f \left( x \right) \cos 2 \pi x$ $f \left( x + 2 \right) \cos 2 \pi \left( x + 2 \right) \mathit{=} f \left( x \right) \cos 2 \pi x$ $\displaystyle \int _{- 1} ^{5} {} f \left( x \right) \cos 2 \pi x dx$ $\displaystyle = \int _{- 1} ^{1} {} f \left( x \right) \cos 2 \pi x dx + \int _{1} ^{3} {f \left( x \right) \cos 2 \pi x dx}$$\displaystyle + \int _{3} ^{5} {} f \left( x \right) \cos 2 \pi x dx$ $\displaystyle = \int _{- 1} ^{1} {} f \left( x \right) \cos 2 \pi x dx$$\displaystyle + \int _{- 1} ^{1} {f \left( x + 2 \right) \cos 2 \pi \left( x + 2 \right) dx}$ $\displaystyle + \int _{- 1} ^{1} {} f \left( x + 4 \right) \cos 2 \pi \left( x + 4 \right) dx$ $\displaystyle = \int _{- 1} ^{1} {} f \left( x \right) \cos 2 \pi x dx + \int _{- 1} ^{1} {f \left( x \right) \cos 2 \pi x dx}$$\displaystyle + \int _{- 1} ^{1} {} f \left( x \right) \cos 2 \pi x dx$ $\displaystyle = 3 \int _{- 1} ^{1} {} f \left( x \right) \cos 2 \pi x dx$ $\displaystyle = 6 \int _{0} ^{1} {} f \left( x \right) \cos 2 \pi x dx$ ㉠, ㉡에 의하여 $\displaystyle \int _{0} ^{1} {} f \left( x \right) \cos 2 \pi x dx$$\displaystyle = \frac{1}{6} \left( \frac{47}{2} - 24 \right)$$\displaystyle = \mathit{-} \frac{1}{12}$ 따라서 $\displaystyle \int _{0} ^{1} {} f' \left( x \right) \sin 2 \pi x dx$$\displaystyle = \left[ \begin{array}{l} f \left( x \right) \sin 2 \pi x \begin{matrix} \\ \end{matrix} \end{array} \right] _{0} ^{1} - 2 \pi \int _{0} ^{1} {} f \left( x \right) \cos 2 \pi x dx$ $\displaystyle = \mathit{-} 2 \pi \int _{0} ^{1} {} f \left( x \right) \cos 2 \pi x dx$$\displaystyle = \frac{\pi}{6}$