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spockmonkey40 2022-07-06 Answered
Antiderivative of x e c x 2
I need to define c in 0 x e c x 2 ,, so that it becomes a probability-mass function (so that it equals 1).
Where do I even begin finding the antiderivative of this? I know the answer will be: e c x 2 2 c .
Trying to use partial integration:
f ( x ) g ( x ) d x = f ( x ) G ( x ) f ( x ) G ( X )
and picking x as my g(x), and e c x 2 as my f(x) I end up with:
f ( x ) = 2 c x f ( x )
G ( X ) = x 2 2
e c x 2 x 2 2 0 2 c x f ( x ) x 2 2 d x
Which simplifies to: e c x 2 x 2 2 0 c x 4 e c x 2 d x, which I find is just a mess.
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Answers (2)

Amir Beck
Answered 2022-07-07 Author has 13 answers
Step 1
The antiderivative is easy. Note that ,since you want the integral to equal 1:
0 x e c x 2 d x = 1 1 2 c 0 2 c x e c x 2 d x = 1 1 2 c [ e c x 2 ] 0 = 1 2 c = 1 c = 1 / 2
Note also that c must be positive , otherwise the integral won't converge.
Step 2
Update: I fixed a major blunder. The quantity in brackets is evaluated as follows:
0 2 c x e c x 2 d x = lim x + e c x 2 + e 0 = 1
(which is the definition of improper integral) and hence we go on to solve the equation.
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glitinosim3
Answered 2022-07-08 Author has 2 answers
Explanation:
For f ( x ) = e c x 2 you have f ( x ) = 2 c x f ( x ), so integration by parts gives x f ( x ) d x = 1 2 c f ( x ) d x = 1 2 c f ( x ) + C.
Now the definite integral.
To have a probability-mass function, c has to be positive (otherwise the limit in infinity is intinite) and then 1 = x f ( x ) d x = 1 2 c f ( x ) | 0 = 1 2 c ( 0 1 ), so c = 1 2 is the only solution.
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