find the average value of the function f(x,y)=e^(-x^2) over the plane region R which is the triangle with vertices (0,0), (1,0) and (1,1)

Answer 1

Please check the function for typo, since
f(x,y)=e^(-x²) is independent of y.

Assuming no typo,
the region R is bounded between x=0 and x=1, and y=0 and y=x.

So the integegration
dy from 0 to x
dx from 0 to 1.

I=∫∫ye^(-x²)dy dx
=∫xe^-x² dx
=1/2-e^(-1)/2

The area of R is 1/2, so
Average value
= I/(1/2)
=(e-1)/e