Random variables and distribution

Question

If X is a continuous random variable whose probability density function is given by
f(x) ={k(5x-2x^2) , 0<= x <= 2<br />            0 ,otherwise}
then p(x>1) is

Answer 1

P(x>1)=1-p(x<1)

integrating the given  function net 0 and 2 gives k(5x^2/2-2x^3/3)

Equate this to 1 to get value ok k

 value of k as 3/14

P(1)=3/14*11/6

=11/28

P(x>1)=1-11/28=17/28

Answer 2

integrate the fun wrt x nd take the limit from 1 to  2 as the value of  function is 0 for larger x.

Comments

I tried that answer given as 17/28 but iam getting different answer

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when u will integrete from 1 to 2 u will get a result which also contain k .The  result u will get is 17/6 k.

 

now since we know that integration of this density fun from -infinte to +infinite should be 1.

integrate f(x) from -infinite to +infinite which should be 1.

so,after integration the eqn will be 14k/3  = 1  ,so k=3/14

now put this k value into our required result = 17k/6  = (17/6 ) * (3/14)  =  17/28