A calculus exercise from a probability distribution

Several calculus exercises that are derived from a probability problem are presented here.

Let \alpha>0 be a constant.

Evaluate \displaystyle \int_0^1 (1-e^{-\alpha z}) \ dz + \int_1^{\infty}(e^\alpha-1) e^{-\alpha z} \ dz.

Evaluate \displaystyle \int_0^1 z \ (1-e^{-\alpha z}) \ dz + \int_1^{\infty} z \ (e^\alpha-1) e^{-\alpha z} \ dz.

Evaluate \displaystyle \int_0^1 z^2 \ (1-e^{-\alpha z}) \ dz + \int_1^{\infty} z^2 \ (e^\alpha-1) e^{-\alpha z} \ dz.

For anyone who is interested in the origin of these integrals, the first exercise is to verify that the following function is a probability density function (pdf). The second and the third exercises are to find the first and second moment of this probability distribution.

\displaystyle f_Z(z)=\left\{\begin{matrix} \displaystyle 1-e^{-\alpha z}&\ \ \ \ \ \ 0 \le z <1 \\{\text{ }}& \\{\text{ }}& \\{\displaystyle e^{-\alpha z}(e^\alpha-1)}&\ \ \ \ \ \ 1 \le z <\infty  \end{matrix}\right.

The above pdf f_Z(z) is the independent sum of an exponential distribution and the uniform distribution U(0,1). So the answers of the integrals can be derived from knowing these probability distributions. Otherwise, they are excellent exercises in calculus.

\displaystyle \int_0^1 (1-e^{-\alpha z}) \ dz + \int_1^{\infty}(e^\alpha-1) e^{-\alpha z} \ dz=1.

\displaystyle \int_0^1 z \ (1-e^{-\alpha z}) \ dz + \int_1^{\infty} z \ (e^\alpha-1) e^{-\alpha z} \ dz=\frac{1}{2}+\frac{1}{\alpha}.

\displaystyle \int_0^1 z^2 \ (1-e^{-\alpha z}) \ dz + \int_1^{\infty} z^2 \ (e^\alpha-1) e^{-\alpha z} \ dz=\frac{1}{3}+\frac{1}{\alpha}+\frac{2}{\alpha^2}.

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