확률밀도함수 기댓값

${\mathrm{P}} ( 0 \leq X \leq 4 )$$=$$\displaystyle \int _{0} ^{4} f ( x ) dx = 1$이어야하므로 $\displaystyle \int _{0} ^{4} f ( x ) dx$$\displaystyle = \int _{0} ^{1} f ( x ) dx + \int _{1} ^{4} f ( x ) dx$ $\displaystyle = \int _{0} ^{1} \frac{1}{2} x dx + \int _{1} ^{4} a ( x - 4 ) dx$ $\displaystyle = \left[ \frac{x ^{2}}{4} \right] _{0} ^{1} + a \left[ \frac{x ^{2}}{2} - 4 x \right] _{1} ^{4}$ $\displaystyle = \frac{1}{4} - \frac{9}{2} a$ $= 1$ ∴ $\displaystyle a = - \frac{1}{6}$ ∴ ${\mathrm{E}} ( X )$$\displaystyle = \int _{0} ^{4} x f ( x ) dx$ $\displaystyle = \int _{0} ^{1} xf ( x ) dx + \int _{1} ^{4} xf ( x ) dx$ $\displaystyle = \int _{0} ^{1} \frac{x ^{2}}{2} dx + \int _{1} ^{4} \frac{4 x - x ^{2}}{6} dx$ $\displaystyle = \left[ \frac{x ^{3}}{6} \right] _{0} ^{1} + \left[ \frac{x ^{2}}{3} - \frac{x ^{3}}{18} \right] _{1} ^{4}$ $\displaystyle = \frac{1}{6} + \frac{3}{2}$$\displaystyle = \frac{5}{3}$ ∴ ${\mathrm{E}} ( 6 X + 5 )$$= 6 \times {\mathrm{E}} ( X ) + 5$$= 10 + 5$$= 15$ [다른 풀이] $f ( x )$가 연속확률변수 $X$의 확률밀도함수이므로 함수 $f ( x )$는 $x = 1$에서 연속이어야 한다. $\displaystyle f ( 1 ) = \lim\limits _{x \rightarrow 1 +} {} f ( x )$에서 $\displaystyle \frac{1}{2} = - 3 a$ ∴ $\displaystyle a = - \frac{1}{6}$ ⋮ (생략)