약학과 1번

1-1[1단계]$\displaystyle \begin{aligned}&\int_0^a x^n(a-x)^n\,dx\\&=\frac1{n+1}\Biggl[x^{n+1}(a-x)^n\Biggr]_0^a+\frac n{n+1}\int_0^a x^{n+1}(a-x)^{n-1}\,dx\\&=\frac n{n+1}\frac{n-1}{n+2}\int_0^a x^{n+2}(a-x)^{n-2}\,dx\\&=\frac n{n+1}\frac{n-1}{n+2}\cdots\frac2{2n-1}\frac1{2n}\int_0^a x^{2n}\,dx\end{aligned}$$\displaystyle \int_0^a x^{2n}\,dx=\frac{a^{2n+1}}{2n+1}$$\displaystyle \therefore\int_0^a x^n(a-x)^n\,dx=\frac{(n!)^2}{(2n+1)!}a^{2n+1}=A$

[2단계]$\displaystyle B=\int_1^a\ln x\,dx=\Biggl[x\ln x\Biggr]_1^a-\int_1^a1\,dx=a\ln a-a+1$$\displaystyle \begin{aligned}&(2n+1)!A=(n!)^2a^{2n+1}B\\&\Rightarrow1=B\Rightarrow a\ln a-a=0\quad\therefore a=e\end{aligned}$닫힌 구간 $\displaystyle [0,\pi]$에서 $\displaystyle e^x\sin x\ge0$이므로, 구하는 넓이는 $\displaystyle \int_0^\pi a^x\sin x\,dx$이다.$\displaystyle \therefore\int_0^\pi e^x\cos x\,dx=\Biggl[e^x\cos x\Biggr]_0^\pi+\int_0^\pi e^x\sin x\,dx=-e^\pi-1+\int_0^\pi e^x\sin x\,dx$$\displaystyle \therefore\int_0^\pi e^x\sin x\,dx=\Biggl[e^x\sin x\Biggr]_0^\pi-\int_0^\pi e^x\cos x\,dx$$\displaystyle 2\int_0^\pi e^x\sin x\,dx=e^\pi+1$$\displaystyle \therefore\text{넓이}=\int_0^\pi e^x\sin x\,dx=\frac{e^\pi+1}{2}$

1-2[1단계]주어진 조건 $\displaystyle f^{\prime\prime}(x)=f^{\prime\prime}(-x)$로부터 $\displaystyle f^{\prime}(x)=f^{\prime}(0)+\int_0^x f^{\prime\prime}(t)\,dt\;\cdots\;(1)$$\displaystyle -x$를 대입하여 $\displaystyle f^{\prime}(-x)=f^{\prime}(0)+\int_0^{-x}f^{\prime\prime}(t)\,dt$$\displaystyle f^{\prime\prime}(x)=f^{\prime\prime}(-x)$임을 알고 있으므로 $\displaystyle t=-s$로 치환하면$\displaystyle f^{\prime}(-x)=f^{\prime}(0)+\int_0^x f^{\prime\prime}(-s)(-ds)=f^{\prime}(0)-\int_0^x f^{\prime\prime}(-s)\,ds=f^{\prime}(0)-\int_0^x f^{\prime\prime}(t)\,dt$즉, $\displaystyle f^{\prime}(-x)=f^{\prime}(0)-\int_0^x f^{\prime\prime}(t)\,dt\;\cdots\;(2)$$\displaystyle f^{\prime}(x)+f^{\prime}(-x)=2f^{\prime}(0)$$\displaystyle \therefore f^{\prime}(x)=2f^{\prime}(0)-f^{\prime}(-x)$$\displaystyle f^{\prime}(0)=0$인 경우 $\displaystyle f^{\prime}(-x)=-f^{\prime}(x)$이다.

[2단계]$\displaystyle \int_{-a}^a f^{\prime}(-x)\,dx=\int_a^{-a}f^{\prime}(t)(-dt)=\int_{-a}^a f^{\prime}(t)\,dt=\int_{-a}^a f^{\prime}(x)\,dx$$\displaystyle \begin{aligned}2\int_{-a}^a f^{\prime}(x)\,dx&=\int_{-a}^a f^{\prime}(x)\,dx+\int_{-a}^a f^{\prime}(-x)\,dx=\int_{-a}^a(f^{\prime}(x)+f^{\prime}(-x))\,dx\\&=\int_{-a}^a2f^{\prime}(0)\,dx\end{aligned}$$\displaystyle \Biggl[2f^{\prime}(0)x\Biggr]_{-a}^a=2f^{\prime}(0)a-2f^{\prime}(0)(-a)=4af^{\prime}(0)\;\cdots\;(3)$$\displaystyle \int_{-a}^a f^{\prime}(x)\,dx=2af^{\prime}(0)$$\displaystyle \therefore$ 조건에서 $\displaystyle f^{\prime}(0)=0\;\cdots\;(4)$$\displaystyle 0=\int_{-a}^a f^{\prime\prime}(x)\,dx=2\int_0^a f^{\prime\prime}(x)\,dx=2\Biggl[f^{\prime}(x)\Biggr]_0^a=2f^{\prime}(a)-2f^{\prime}(0)$$\displaystyle \therefore f^{\prime}(a)=f^{\prime}(0)$$\displaystyle f^{\prime}(-a)=f^{\prime}(a)=f^{\prime}(0)$이고,(4)에서 $\displaystyle \int_{-a}^a f^{\prime}(x)\,dx=0$$\displaystyle \begin{gathered}\int_{-a}^a f^{\prime}(x)\,dx=2f^{\prime}(0)=0\\\therefore f^{\prime}(0)=0\\\therefore f^{\prime}(a)=0,\;f^{\prime}(-a)=0\end{gathered}$Correction. 이 줄의 $\displaystyle 2f^{\prime}(0)$에는 계수 $\displaystyle a$가 빠져 있다. 그러나 원칙에 따라 원문 표기를 그대로 실었다.이고, $\displaystyle f^{\prime\prime}(a)\ne0$이므로$\displaystyle \therefore f(x)$는 $\displaystyle a$, $\displaystyle -a$에서 극값을 갖는다.