The value of constant "c" for which the following function is a density function:
[tex]f(x) = \frac{c}{x^2 + 1}[/tex], where [tex]-\infty < x < \infty[/tex]
A. 0
B. [tex] \frac{1}{\pi}[/tex]
C. [tex]\pi[/tex]
D. 1
Answer (1)
To determine the value of the constant c for which the given function f ( x ) = x 2 + 1 c is a probability density function, we need to ensure that the integral of f ( x ) over the entire real line is equal to 1. This is because the area under the curve of a probability density function must equal 1.
The function is given by: f ( x ) = x 2 + 1 c , − ∞ < x < ∞
We need to solve: ∫ − ∞ ∞ x 2 + 1 c d x = 1
The integral ∫ − ∞ ∞ x 2 + 1 1 d x is known to equal π . This integral is a standard result, stemming from the fact that this function represents a scaled version of the inverse tangent function's derivative. Therefore, we can plug this into our equation:
c ⋅ ∫ − ∞ ∞ x 2 + 1 1 d x = c ⋅ π
Setting this equal to 1, we have: c ⋅ π = 1
Solving for c :
c = π 1
Thus, the correct value for c is π 1 . The choice that corresponds to this result is (B) \frac{1}{\pi} .
Therefore, the value of constant c for which the function f ( x ) becomes a density function is π 1 .
