Calculate the expected value of a random variable X if f(x) = x – 5⁄2 for 0<x<1 and 2x for 1<x<2 and 0 otherwise.
E(x)=∫−∞+∞xf(x)dx=∫01x(x−52)dx+∫12x⋅2xdx=∫01(x2−5x2)dx+∫122x2dx=x33∣01−5x24∣01+2x33∣12=13−54+2(8−1)3=154E(x) = \int\limits_{ - \infty }^{ + \infty } {xf(x)dx = } \int\limits_0^1 {x\left( {x - \frac{5}{2}} \right)} dx + \int\limits_1^2 {x \cdot 2xdx} = \int\limits_0^1 {\left( {{x^2} - \frac{{5x}}{2}} \right)} dx + \int\limits_1^2 {2{x^2}} dx = \left. {\frac{{{x^3}}}{3}} \right|_0^1 - \left. {\frac{{5{x^2}}}{4}} \right|_0^1 + \left. {\frac{{2{x^3}}}{3}} \right|_1^2 = \frac{1}{3} - \frac{5}{4} + \frac{{2(8 - 1)}}{3} = \frac{{15}}{4}E(x)=−∞∫+∞xf(x)dx=0∫1x(x−25)dx+1∫2x⋅2xdx=0∫1(x2−25x)dx+1∫22x2dx=3x3∣∣01−45x2∣∣01+32x3∣∣12=31−45+32(8−1)=415
Answer: E(x)=154E(x) = \frac{{15}}{4}E(x)=415
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