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प्रश्न
The probability density function of X is given by
`f(x) = {{:(kx"e"^(-2x), "for" x > 0),(0, "for" x ≤ 0):}`
Find the value of k
उत्तर
`f(x) = {{:(kx"e"^(-2x), "for" x > 0),(0, "for" x ≤ 0):}`
Since, f(x) is a Probability density function
`int_oo^oo f(x) "d"x` = 1
i.e., `int_oo^0 0 "d"x + k int_0^oo x"e"^(-2x) "d"x` = 1
`k int_0^oo x"e"^(-2x) "d"x` = 1
Using, integration by parts method
Let u = x
`int "dv" = int "e"^(-2x) "d"x`
di = dx
v = `"e"^(-2x)/(-2)`
`int "u" "dv" = "uv" - int "v" "du"`
`int x"e"^(-2x) "d"x = (x"e"^(-2x))/(-2) + 1/2 int "e"^(-2x) "d"x`
= `- (x"e"^(-2x))/2 - "e"^(-2x)/4`
Now `k int_0^oo x"e"^(-2x) "d"x` = 1
`k[- (x"e"^(-2x))/2 - "e"^(-2x)/4]_0^oo` = 1
`k[0 + (0 + 1/4)]` = 1
⇒ `k(1/4)` = 1
k = 4
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