Answer:
[tex]\boxed{c=\frac{2}{15}}[/tex]
Step-by-step explanation:
The region R of integration is depicted on the picture attached.
To find the area under the surface f(x,y) = c on that region we must evaluate
[tex]\iint_R \ f(x,y) \,dx\,dy=\iint_R \ c \,dx\,dy=c\iint_R \ \,dx\,dy[/tex]
But, by watching the region R we can see
[tex]c\iint_R \ \,dx\,dy=c\left ( \int_{0}^{1}\int_{0}^{x+1}dxdy+\int_{1}^{4}\int_{x-1}^{x+1}dxdy \right )[/tex]
Evaluating the integrals
[tex]\int_{0}^{1}\int_{0}^{x+1}dxdy=\int_{0}^{1}\left ( \int_{0}^{x+1} dy\right )dx=\int_{0}^{1}(x+1)dx=\\\ \int_{0}^{1}xdx+\int_{0}^{1}dx=\frac{3}{2}[/tex]
[tex]\int_{1}^{4}\int_{x-1}^{x+1}dxdy \right=\int_{1}^{4}\left ( \int_{x-1}^{x+1}dy \right )dx=\int_{1}^{4}2dx=6[/tex]
And we have that the area under the surface is
[tex]c(\frac{3}{2}+6)=\frac{15}{2}c[/tex]
Since f is a probability density function
[tex]\frac{15}{2}c=1[/tex]
and we must have
[tex]\boxed{c=\frac{2}{15}}[/tex]
