(a) f ( x ) \displaystyle f ( x ) f ( x ) ∫ 0 a ( t + 1 ) e − x d x = [ − ( t + 1 ) e − x ] 0 a = ( t + 1 ) ( 1 − e − a ) = 1 \displaystyle \int_{0}^{a} {( t + 1 ) e^{- x} d x} = \Biggl[ - ( t + 1 ) e^{- x} \Biggr]_{0}^{a}= ( t + 1 ) ( 1 - e^{- a} ) = 1 ∫ 0 a ( t + 1 ) e − x d x = [ − ( t + 1 ) e − x ] 0 a = ( t + 1 ) ( 1 − e − a ) = 1 이다. 따라서 ( t + 1 ) ( 1 − e − a ) = 1 ⇔ ( 1 − e − a ) = 1 t + 1 ⇔ e − a = t t + 1 ⇔ e a = t + 1 t \displaystyle (t+1)(1-e^{-a})=1\Leftrightarrow(1-e^{-a})=\frac1{t+1}\Leftrightarrow e^{-a}=\frac t{t+1}\Leftrightarrow e^a=\frac{t+1}t ( t + 1 ) ( 1 − e − a ) = 1 ⇔ ( 1 − e − a ) = t + 1 1 ⇔ e − a = t + 1 t ⇔ e a = t t + 1 이므로 a = ln ( 1 + 1 t ) \displaystyle a = \ln ( 1 + \frac{{1}}{t} ) a = ln ( 1 + t 1 )
(b) E ( X 2 ) = ∫ 0 a x 2 f ( x ) d x \displaystyle \mathrm E(X^2)=\int_0^a x^2f(x)\,dx E ( X 2 ) = ∫ 0 a x 2 f ( x ) d x E ( X 2 ) = ∫ 0 ln ( 1 + 1 t ) x 2 ⋅ ( t + 1 ) e − x d x = ( t + 1 ) ∫ 0 ln ( 1 + 1 t ) x 2 e − x d x = ( t + 1 ) [ − x 2 e − x ] 0 ln ( 1 + 1 t ) + ( t + 1 ) ∫ 0 ln ( 1 + 1 t ) 2 x e − x d x = ( t + 1 ) ( − t t + 1 [ ln ( 1 + 1 t ) ] 2 ) + 2 ( t + 1 ) ∫ 0 ln ( 1 + 1 t ) x e − x d x = − t [ ln ( 1 + 1 t ) ] 2 + 2 ( t + 1 ) [ − x e − x ] 0 ln ( 1 + 1 t ) + 2 ( t + 1 ) ∫ 0 ln ( 1 + 1 t ) e − x d x = − t [ ln ( 1 + 1 t ) ] 2 − 2 t ln ( 1 + 1 t ) + 2 ( t + 1 ) ∫ 0 ln ( 1 + 1 t ) e − x d x = − t [ ln ( 1 + 1 t ) ] 2 − 2 t ln ( 1 + 1 t ) + 2 \displaystyle \begin{aligned}\mathrm E(X^2)&=\int_0^{\ln(1+\frac1t)}x^2\cdot(t+1)e^{-x}\,dx=(t+1)\int_0^{\ln(1+\frac1t)}x^2e^{-x}\,dx\\&=(t+1)\Biggl[-x^2e^{-x}\Biggr]_0^{\ln(1+\frac1t)}+(t+1)\int_0^{\ln(1+\frac1t)}2xe^{-x}\,dx\\&=(t+1)\left(-\frac t{t+1}\left[\ln\left(1+\frac1t\right)\right]^2\right)+2(t+1)\int_0^{\ln(1+\frac1t)}xe^{-x}\,dx\\&=-t\left[\ln\left(1+\frac1t\right)\right]^2+2(t+1)\Biggl[-xe^{-x}\Biggr]_0^{\ln(1+\frac1t)}+2(t+1)\int_0^{\ln(1+\frac1t)}e^{-x}\,dx\\&=-t\left[\ln\left(1+\frac1t\right)\right]^2-2t\ln\left(1+\frac1t\right)+2(t+1)\int_0^{\ln(1+\frac1t)}e^{-x}\,dx\\&=-t\left[\ln\left(1+\frac1t\right)\right]^2-2t\ln\left(1+\frac1t\right)+2\end{aligned} E ( X 2 ) = ∫ 0 l n ( 1 + t 1 ) x 2 ⋅ ( t + 1 ) e − x d x = ( t + 1 ) ∫ 0 l n ( 1 + t 1 ) x 2 e − x d x = ( t + 1 ) [ − x 2 e − x ] 0 l n ( 1 + t 1 ) + ( t + 1 ) ∫ 0 l n ( 1 + t 1 ) 2 x e − x d x = ( t + 1 ) ( − t + 1 t [ ln ( 1 + t 1 ) ] 2 ) + 2 ( t + 1 ) ∫ 0 l n ( 1 + t 1 ) x e − x d x = − t [ ln ( 1 + t 1 ) ] 2 + 2 ( t + 1 ) [ − x e − x ] 0 l n ( 1 + t 1 ) + 2 ( t + 1 ) ∫ 0 l n ( 1 + t 1 ) e − x d x = − t [ ln ( 1 + t 1 ) ] 2 − 2 t ln ( 1 + t 1 ) + 2 ( t + 1 ) ∫ 0 l n ( 1 + t 1 ) e − x d x = − t [ ln ( 1 + t 1 ) ] 2 − 2 t ln ( 1 + t 1 ) + 2 이다. 또한 lim t → ∞ E ( X 2 ) \displaystyle \lim_{t\to\infty}\mathrm E(X^2) t → ∞ lim E ( X 2 ) lim t → ∞ E ( X 2 ) = lim t → ∞ ( − t [ ln ( 1 + 1 t ) ] 2 − 2 t ln ( 1 + 1 t ) + 2 ) = lim t → ∞ ( − 1 t [ ln ( 1 + 1 t ) t ] 2 − 2 ln ( 1 + 1 t ) t + 2 ) = 0 − 2 ln e + 2 = 0 \displaystyle \begin{aligned}\lim_{t\to\infty}\mathrm E(X^2)&=\lim_{t\to\infty}\left(-t\left[\ln\left(1+\frac1t\right)\right]^2-2t\ln\left(1+\frac1t\right)+2\right)\\&=\lim_{t\to\infty}\left(-\frac1t\left[\ln\left(1+\frac1t\right)^t\right]^2-2\ln\left(1+\frac1t\right)^t+2\right)=0-2\ln e+2\\&=0\end{aligned} t → ∞ lim E ( X 2 ) = t → ∞ lim ( − t [ ln ( 1 + t 1 ) ] 2 − 2 t ln ( 1 + t 1 ) + 2 ) = t → ∞ lim − t 1 [ ln ( 1 + t 1 ) t ] 2 − 2 ln ( 1 + t 1 ) t + 2 = 0 − 2 ln e + 2 = 0
QtA. 왜 lim t → ∞ E ( X 2 ) = 0 \displaystyle\lim_{t\to\infty}\mathrm E(X^2)=0 t → ∞ lim E ( X 2 ) = 0 이 문제에서 X \displaystyle X X 0 ≤ X ≤ a = ln ( 1 + 1 t ) \displaystyle 0\le X\le a=\ln\left(1+\frac1t\right) 0 ≤ X ≤ a = ln ( 1 + t 1 ) 이다. t \displaystyle t t a \displaystyle a a X \displaystyle X X X 2 \displaystyle X^2 X 2 0 ≤ E ( X 2 ) ≤ a 2 = { ln ( 1 + 1 t ) } 2 ⟶ 0 \displaystyle 0\le \mathrm E(X^2)\le a^2=\left\{\ln\left(1+\frac1t\right)\right\}^2\longrightarrow0 0 ≤ E ( X 2 ) ≤ a 2 = { ln ( 1 + t 1 ) } 2 ⟶ 0 이므로 구하는 극한값은 0 \displaystyle 0 0
수리논술, 이게 다다.