자연계열 2번

[2-1 풀이]

$\displaystyle \ln \frac{{n}}{\sqrt[n]{( n + 4 ) \times ( n + 8 ) \times ( n + 12 ) \times \cdots \times ( n + 4 n )}}$

$\displaystyle = - \frac{{1}}{4} 4 \ln \frac{{\sqrt[n]{( n + 4 ) \times ( n + 8 ) \times ( n + 12 ) \times \cdots \times ( n + 4 n )}}}{n}$

$\displaystyle = - \frac{1}{4} \times {\frac{{4}}{n} \left\{ \ln \left( 1 + \frac{{4}}{n} \right) + \ln \left( 1 + \frac{{8}}{n} \right) + \ln \left( 1 + \frac{{12}}{n} \right) + \cdots + \ln \left( 1 + \frac{{4 n}}{n} \right) \right\}}$

$\displaystyle = - \frac{1}{4} \times {\frac{{4}}{n} \sum_{k = 1}^{n} \ln \left( 1 + \frac{{4 k}}{n} \right)}$

따라서

$\displaystyle \lim_{n \to \infty} {\ln \left\{ \frac{{n}}{\sqrt[n]{( n + 4 ) \times ( n + 8 ) \times ( n + 12 ) \times \cdots \times ( n + 4 n )}} \right\}}$

$\displaystyle = - \frac{1}{4} \lim_{n \to \infty} {\frac{{4}}{n} \sum_{k = 1}^{n} \ln \left( 1 + \frac{{4 k}}{n} \right)}$

$\displaystyle = - \frac{1}{4} \int_{0}^{4} {\ln ( 1 + x ) d x}$

$\displaystyle = - \frac{{1}}{4} \int_{1}^{5} {\ln t} d t= - \frac{1}{4} \Biggl[ t \ln t \Biggr]_{1}^{5} + \frac{1}{4} \int_{1}^{5} 1 d t= \frac{{4 - 5 \ln 5}}{4}$

[2-2 풀이]

$\displaystyle f^{\prime} ( x ) = \frac{{1}}{2} \left( \frac{{1}}{3 + x} - \frac{{1}}{3 - x} \right) = - \frac{{x}}{( 3 + x ) ( 3 - x )}$이므로

$\displaystyle f ( - 2 ) > 0 , f ( 2 ) > 0$ 이고 $\displaystyle f ( - x ) = f ( x )$이므로 $\displaystyle y$축에 대칭인 그래프이다. 그러므로 구하고자 하는 넓이는

$\displaystyle S= 2 \int_{0}^{2} {f ( x )} d x= 2 \int_{0}^{2} {\ln \sqrt{( 3 + x ) ( 3 - x )}} d x$$\displaystyle = \int_{0}^{2} {\left\{ \ln ( 3 + x ) + \ln ( 3 - x ) \right\} d x}= \int_{0}^{2} {\ln ( 3 + x )} d x + \int_{0}^{2} {\ln ( 3 - x )} d x$$\displaystyle = \int_{3}^{5} {\ln t} d t - \int_{3}^{1} {\ln s} d s= \int_{3}^{5} {\ln t} d t + \int_{1}^{3} {\ln s} d s$$\displaystyle = \int_{1}^{5} {\ln t} d t$

따라서 문제 1의 결과를 이용하면 $\displaystyle S = \int_{1}^{5} {\ln t} d t= 5 \ln 5 - 4$