Answer :
The answer will be a = 1.6 and b = 11 /12
What is integration?
Integration is defined as the summing up of the small parts to find the total value.
The solution of the given question is as follows:-
The joint pdf by x and y
1 < x <2
0 < x <1
In point a:
[tex]K\int^1_{y=0} \int_{x=1}^2( 2y-\dfrac{x}{4}dxdy=1\\\\\\K\int _{y=0}^12y-\dfrac{x^2}{8}\int_1^2dy=1\\\\\\K\ y^2\int_{y=0}^1-\dfrac{3}{8}=1\\\\\\K(1-\dfrac{3}{8})=1\\\\\\K=\dfrac{8}{5}[/tex]
K = 1.6
In point b:
In the original pdf of x
[tex]f(x)=K\int_{y=0}^1(2y-\dfrac{x}{4})dy=k\\\\\\K\int_1^2(x-\dfrac{x^2}{4})dx=K\\\\\\K(\dfrac{x^2}{2}-\dfrac{x^3}{12})_1^2=K\\\\\\K(\dfrac{1}{2}\times 3)-\dfrac{1}{12}\times 7)=K\\\\\dfrac{3}{2}-\dfrac{7}{12}\\\\\\=\dfrac{11}{12}[/tex]
Therefore the answer will be a = 1.6 and b = 11 /12
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