자연계열 3번

3-1. [풀이]$\displaystyle x=\pi-t$로 놓으면 $\displaystyle dx=-dt$이므로$\displaystyle \begin{aligned}\int_0^\pi xf(\sin x)dx&=\int_\pi^0(\pi-t)f(\sin(\pi-t))(-dt)\\&=\int_0^\pi(\pi-t)f(\sin t)dt\\&=\int_0^\pi\pi f(\sin t)dt-\int_0^\pi tf(\sin t)dt\end{aligned}$

따라서 $\displaystyle 2\int_0^\pi xf(\sin x)dx=\pi\int_0^\pi f(\sin x)dx$이므로$\displaystyle \int_0^\pi xf(\sin x)dx=\frac\pi2\int_0^\pi f(\sin x)dx$

3-2. [풀이]$\displaystyle \frac{dx}{dy}=g^{\prime}(y)$이므로,$\displaystyle \begin{aligned}\int_a^b f(x)dx&=\int_{f(a)}^{f(b)}yg^{\prime}(y)dy\\&=\Biggl[yg(y)\Biggr]_{f(a)}^{f(b)}-\int_{f(a)}^{f(b)}g(y)dy\\&=bf(b)-af(a)-\int_{f(a)}^{f(b)}g(y)dy\end{aligned}$

3-3. [풀이]$\displaystyle y=\ln(1+\sqrt{x})$라 할 때, $\displaystyle 1+\sqrt{x}=e^y$, $\displaystyle x=(e^y-1)^2$이므로, $\displaystyle f(x)=\ln(1+\sqrt{x})$의 역함수는$\displaystyle g(x)=(e^x-1)^2$이다.

한편, $\displaystyle F(t)=\int_0^t\ln(1+\sqrt{x})dx-at$이므로 $\displaystyle F^{\prime}(t)=\ln(1+\sqrt{t})-a$이고, $\displaystyle F^{\prime\prime}(t)>0$이다.따라서 $\displaystyle t=(e^a-1)^2$일 때, 극소이면서 최소이다.

이때 $\displaystyle t$값은 $\displaystyle g(a)$와 같으므로, 제시문 <다>를 이용하면,$\displaystyle \begin{aligned}F(g(a))&=\int_0^{g(a)}\ln(1+\sqrt{x})dx-ag(a)=-\int_0^a g(y)dy\\&=-\int_0^a(e^y-1)^2dy\\&=-\Biggl[\frac{e^{2y}}2-2e^y+y\Biggr]_0^a\\&=-\frac{e^{2a}}2+2e^a-a-\frac32\end{aligned}$따라서 $\displaystyle F(t)$의 최솟값은 $\displaystyle -\frac{e^{2a}}2+2e^a-a-\frac32$이다.