[dispmath]E(X)=\int\limits_{-\infty}^\infty xf(x)\,\mathrm dx=\int\limits_2^6x\cdot\left(\frac{x}{8}-\frac{1}{4}\right)\,\mathrm dx=\int\limits_2^6\left(\frac{x^2}{8}-\frac{x}{4}\right)\,\mathrm dx=\left.\left(\frac{x^3}{16}-\frac{x^2}{4}\right)\right|_2^6=\\
=\left(\frac{6^3}{16}-\frac{6^2}{4}\right)-\left(\frac{2^3}{16}-\frac{2^2}{4}\right)=\left(\frac{216}{16}-\frac{36}{4}\right)-\left(\frac{8}{16}-\frac{4}{4}\right)=\\
=\left(\frac{216}{16}-\frac{144}{16}\right)-\left(\frac{8}{16}-\frac{16}{16}\right)=\frac{72}{16}-\left(-\frac{8}{16}\right)=\frac{72}{16}+\frac{8}{16}=\frac{80}{16}=5[/dispmath] Ovako nešto?