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To find the probability that the system functions for at least 5 months, we need to integrate the given density function from 5 months to infinity. The probability can be calculated as follows:
$P(X\ge 5)={\int }_5^{\infty }f(x),dx$
Substituting the given density function into the integral, we have:
$P(X\ge 5)={\int }_5^{\infty }Cx{e}^{-x/2},dx$
To solve this integral, we can use integration by parts. Let's denote $u=x$ and $dv=C{e}^{-x/2}dx$. Then, $du=dx$ and $v=-2C{e}^{-x/2}$. Applying integration by parts, we get:
$P(X\ge 5)={[-2Cx{e}^{-x/2}]}_5^{\infty }-{\int }_5^{\infty }(-2C{e}^{-x/2}),dx$
Evaluating the definite integral, we have:
$P(X\ge 5)={[-2Cx{e}^{-x/2}]}_5^{\infty }+2C{\int }_5^{\infty }{e}^{-x/2},dx$
Next, we can evaluate the limits of the first term:
${[-2Cx{e}^{-x/2}]}_5^{\infty }={lim}_{x\to \infty }(-2Cx{e}^{-x/2})-(-2C(5)e^{-5/2})$
Since $e^{-\infty }=0$, the first term evaluates to $0-(-2C(5)e^{-5/2})=10C{e}^{-5/2}$. Therefore, the probability expression becomes:
$P(X\ge 5)=10C{e}^{-5/2}+2C{\int }_5^{\infty }{e}^{-x/2},dx$
To evaluate the remaining integral, we can use the exponential integral function, denoted as $\text{Ei}(x)$, which is defined as:
$\text{Ei}(x)=-{\int }_{-x}^{\infty }\frac{e^{-t}}{t},dt$
Applying the exponential integral function to our integral, we get:
${\int }_5^{\infty }{e}^{-x/2}dx=-2\text{Ei}(-\frac{x}{2}){|}_5^{\infty }$
Substituting this back into the probability expression, we have:
$P(X\ge 5)=10C{e}^{-5/2}+2C(-2\text{Ei}(-\frac{x}{2}){|}_5^{\infty })$
Simplifying further, we obtain the final expression for the probability:
$P(X\ge 5)=10C{e}^{-5/2}-4C\text{Ei}(-\frac{5}{2})$
Thus, the probability that the system functions for at least 5 months is $10C{e}^{-5/2}-4C\text{Ei}(-\frac{5}{2})$

To solve the given problem, we need to find the probability that the system functions for at least 5 months, given the density function $(f(x)=Cx{e}^{-x/2})$.
First, we need to determine the value of the constant (C) in the density function. To do this, we'll integrate the density function over its entire range and set the result equal to 1, since the total probability must be equal to 1.
${\int }_{-\infty }^{\infty }f(x)dx=1$
However, since the density function is defined as zero for $x\le 0$, we can integrate from 0 to infinity instead.
${\int }_0^{\infty }Cx{e}^{-x/2},dx=1$
Now, we can solve this integral to find the value of $C$. Let's perform the integration:
${\int }_0^{\infty }Cx{e}^{-x/2},dx$
To evaluate this integral, we can use integration by parts. Let's define $u=x$ and $dv=C{e}^{-x/2}dx$. Then, we have $du=dx$ and $v=-2C{e}^{-x/2}$.
Using the integration by parts formula:
$\int udv=uv-\int v,du$
we can rewrite the integral as:
${-2Cx{e}^{-x/2}|}_0^{\infty }-{\int }_0^{\infty }-2C{e}^{-x/2},dx$
Evaluating the limits of the first term, we get:
${lim}_{x\to \infty }(-2Cx{e}^{-x/2})-(-2C·0·e^{0/2})$
Since $e^{-\infty }=0$ and $e^0=1$, this simplifies to:
$0+2C=1$
Therefore, $C=\frac{1}{2}$.
Now that we have determined the value of $C$, we can find the probability that the system functions for at least 5 months. This probability corresponds to the integral of the density function from 5 to infinity:
$P(X\ge 5)={\int }_5^{\infty }\frac{1}{2}x{e}^{-x/2},dx$
To solve this integral, we can once again use integration by parts. Let's define $u=x$ and $dv=\frac{1}{2}{e}^{-x/2},dx$. Then, $du=dx$ and $v=-e^{-x/2}$.
Applying the integration by parts formula:
$\int udv=uv-\int v,du$
we can rewrite the integral as:
${-x{e}^{-x/2}|}_5^{\infty }+{\int }_5^{\infty }{e}^{-x/2},dx$
Evaluating the limits of the first term, we have:
Since $e^{-\infty }=0$, the first term simplifies to $0+5e^{-5/2}$.
Now, let's evaluate the integral:
${\int }_5^{\infty }{e}^{-x/2},dx$
To integrate this, we can make a substitution. Let $u=-\frac{x}{2}$, then $du=-\frac{1}{2},dx$. When $x=5$, we have $u=-\frac{5}{2}$, and when $x=\infty$, $u=-\infty$.
Substituting the values, the integral becomes:
${\int }_{-\frac{5}{2}}^{-\infty }{e}^u·(-2)du=-2{\int }_{-\frac{5}{2}}^{-\infty }{e}^u,du$
Integrating $e^u$ gives us:
${-2e^u|}_{-\frac{5}{2}}^{-\infty }=-2(e^{-\frac{5}{2}}-e^{-\infty })$
Since $e^{-\infty }=0$, the integral simplifies to $-2e^{-\frac{5}{2}}$.
Combining this with the first term, we have:
$5e^{-\frac{5}{2}}-2e^{-\frac{5}{2}}=3e^{-\frac{5}{2}}$
Therefore, the probability that the system functions for at least 5 months is $3e^{-\frac{5}{2}}$.
In summary, the probability that the system functions for at least 5 months is $\boxed{3e^{-\frac{5}{2}}}$.