The $ \int_Tf(x)g(y)dxdy$ equals?

Asked May 16 '23 at 21:04

Active May 16 '23 at 21:04

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Given two real-valued functions f(x) and g(x) both defined by definition

x if x $\in [0,1] $,

0 otherwise.

Let T denote the triangle in the real plane bounded by x axis, y axis and line x+y=2. The $ \int_Tf(x)g(y)dxdy$ equals?

I think it should be simply $\int_0^1 xdx \int_0^1 ydy=\frac{1}{4}$ I want to know if the answer is correct?

asked May 16 '23 at 21:04

Vikas Sharma

Why do you introduce the function $g$? Why not keep it simple and integrate $f(x)f(y)$?? – David G. Stork May 16 '23 at 21:09
1

You are integrating over the unit square; you only want the lower-left half. – ncmathsadist May 16 '23 at 21:11
1

Your answer is correct. – David G. Stork May 16 '23 at 21:18
@ncmathsadist but the square lie within the triangular region. – Vikas Sharma May 16 '23 at 21:20

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