# If the probability density function of #X# is #f(x) = (1+ax) /2, -1 le x le 1, -1 le a le 1# then the expectation of #X# is? (a) #6/a# (b) #a /3# (c) #a /2# (d) #3/a#

##### 2 Answers

#### Explanation:

for a pdf

# E(X) =a/3#

making **(b)** the correct answer

#### Explanation:

By definition if

# E(X) =int_(-oo)^(oo) \ xf(x) \ dx #

So given that

# E(X) = int_(-1)^(1) \ x \ ((1+ax)/2) \ dx #

# \ \ \ \ \ \ \ \ = 1/2 \ int_(-1)^(1) \ x (1+ax) \ dx #

# \ \ \ \ \ \ \ \ = 1/2 \ int_(-1)^(1) \ x+ax^2 \ dx #

# \ \ \ \ \ \ \ \ = 1/2 [ x^2/2+(ax^3)/3 ]_(-1)^(1)#

# \ \ \ \ \ \ \ \ = 1/2 {(1/2+a/3)-(1/2-a/3)}#

# \ \ \ \ \ \ \ \ = 1/2 (1/2+a/3-1/2+a/3)#

# \ \ \ \ \ \ \ \ = 1/2 (2a)/3#

# \ \ \ \ \ \ \ \ = a/3#