Why Must the Endpoints of PDF Functions Match at Boundaries?

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thereddevils
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Continuous random variable X is defined in the interval 0 to 4, with

P(X>x)= 1- ax , 0<=x<=3

= b - 1/2 x , 3<x<=4

with a and b as constants. Find a and b.

So the area under the pdf is 1, then i integrated both functions and set up my first equation.

Next, it seems that the endpoints of the functions are equal at x=3. Why is it so? Must a pdf be continuous? I thought its properties are only f(x)>=0 and the area under it is 1.
 
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thereddevils said:
Continuous random variable X is defined in the interval 0 to 4, with

P(X>x)= 1- ax , 0<=x<=3

= b - 1/2 x , 3<x<=4

with a and b as constants. Find a and b.

So the area under the pdf is 1, then i integrated both functions and set up my first equation.

Next, it seems that the endpoints of the functions are equal at x=3. Why is it so? Must a pdf be continuous? I thought its properties are only f(x)>=0 and the area under it is 1.

Umm, how does P(X>x) define a pdf?
 
bpet said:
Umm, how does P(X>x) define a pdf?

It's as written in the original question but nevermind, take it as f(x).