<s>[INST] Compute the mean for the $f(x)=\dfrac {3}{2}\sqrt {x-1},\;1 \leq x \leq 2$ using probability distribution. How I compute the mean for the $$f(x)=\dfrac {3}{2}\sqrt {x-1},\;1 \leq x \leq 2$$ using  probability distribution.
My attempt$$\mu_x=E(X)=\int_{-\infty}^{\infty}xf(x)dx\\= \int\limits_{1}^{2}\,x\cdot\dfrac {3}{2}\sqrt {x-1}\;dx\\= \dfrac {3}{2} \left [ \int\limits_1^2x\sqrt {x-1}\,dx \right]$$
Now, don't understand what to do.  Thanks in advance.
 [/INST] Now, use integration by parts formula$$\displaystyle\int u\; dv=u\;v-\int v\;du\\\text{Let }u=x,\;\;dv=\sqrt{x-1} \; dx\\du=dx,\;\;v=\dfrac{2}{3}(x-1)^{3/2}\\ = \dfrac{3}{2} \left[x\cdot\dfrac{2}{3}(x-1)^{3/2} - \int \dfrac{2}{3}(x-1)^{3/2} \,dx\right]^2_1\\ =\left[ \dfrac{3}{2} \left[x\cdot\dfrac{2}{3}(x-1)^{3/2} - \dfrac{2}{3}×\dfrac{2}{5} (x-1)^{5/2} \,\right]\right]^2_1\\= \left[x(x-1)^{3/2} - \dfrac{2}{5} (x-1)^{5/2}\right]^2_1\\ = 2- \dfrac{2}{5}\\\mu=\dfrac{8}{5} = 1.6 $$
go there for better explanation on probability distribution.
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