## Question: Find the value of c such that

$f(x)= \begin{cases} \frac {x} {6} + c & 0\le x \le 3 \\ 0 & \mbox{elsewhere} \end{cases}$

is a p.d.f. Also find $P(1\le x \le 2)$

# Solution:

$f(x)\le 0 if c\ge 0$; Also we must have

$\int\limits_{-infty}^{infty}f(x)dx=1$

i.e., $\int\limits_{0}^{3} ( \frac {x} {6} + c)dx=1$

On integrating this we get,

$\frac {3} {4} + 3c=1$

Therefore,

$c=\frac {1} {12}$

Now,

$P(1\le x\le 2)=\int\limits_{1}^{2} f(x) dx$

$=\int\limits_{1}^{2}(\frac {x} {6} + \frac {1} {12}) dx$

Which on further integration leads to,

$\frac {1} {12} [(4+2)-(1+1)]=\frac {1} {3}$

Thus,

$P(1\le x \le 2)= \frac {1} {3}$

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