1. Sendo X uma v.a. contínua então mostra-se que, \(\int_{-\infty }^{+\infty }f(x)dx=1\), pois, \(=\int_{-1}^{2}\frac{x^{2}}{3}dx=\frac{1}{3}\int_{-1}^{2}x^{2}dx=\frac{1}{3}\left [ \frac{x^{3}}{3} \right ]_{-1}^{2}=\frac{1}{9}\left [ 8+1 \right ]=1\)
2. \(P(0<X<1)=P(X<1)-P(X\leqslant 0)=\int_{0}^{1}\frac{x^{2}}{3}dx=\frac{1}{9}\left [ x^{3} \right ]_{0}^{1}=\frac{1}{9}\)
3. Pela def. de valor médio de uma v.a., \(E(X)=\int_{-1}^{2}x\frac{x^{2}}{3}dx=\int_{-1}^{2}\frac{x^{3}}{3}dx=\frac{1}{3}\left [ \frac{x^{4}}{4} \right ]_{-1}^{2}=\frac{1}{12}\left [ 16-1 \right ]=\frac{15}{12}\)
e pela def. de variância de uma v.a.,
\(V(X)=E(X^{2})-E^{2}(X)=\int_{-1}^{2}x^{2}\frac{x^{2}}{3}dx-\left ( \frac{15}{12} \right )^{2}=\frac{1}{3}\int_{-1}^{2}x^{4}dx-\left ( \frac{15}{12} \right )^{2}=\frac{1}{3}\left [ \frac{x^{5}}{5} \right ]_{-1}^{2}-\left ( \frac{15}{12} \right )^{2}=\frac{1}{15}\left [ 32+1 \right ]-\left ( \frac{15}{12} \right )^{2}=\left (\frac{33}{15}\right )-\left ( \frac{15}{12} \right )^{2}\)
Bom estudo
