Suppose $f:[0,1] \rightarrow \mathbb{R}$ is continuous. What is the value of $\int_{0}^{1} \int_{x}^{1-x} f(y) d y d x ?$ Use Fubini's theorem 
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*Suppose $f:[0,1] \rightarrow \mathbb{R}$ is continuous. What is the value of $\int_{0}^{1} \int_{x}^{1-x} f(y) d y d x ?$ Again, do not forget to justify any use of Fubini's Theorem.


I evaluated by the online calculator that $\int_{0}^{1} \int_{x}^{1-x} f(y) d y d x=0.$
My attempt. $\{0\leq x\leq 1, x\leq y\leq 1-x\}$ then $\{0\leq y\leq 1, 1-y\leq x\leq y\}$. Hence the integral is:
$$\int_{0}^{1} \int_{1-y}^{y} f(y) d x d y$$ 
but when I evaluate the integral, I didnt get $0$, can you help?
But When you use Fubini's thereom to evaluated the integral, may you say how can we change boundeds of integrals?

Make the variable change $u=1-x$, $x=0, u=1, x=1,u=0, du=-dx$ $\int_0^1\int_x^{1-x}f(y)dydx$
$=\int_1^0\int_{1-u}^uf(y)dy(-du)=\int_1^0\int_u^{1-u}f(y)dydu=-\int_0^1\int_u^{1-u}f(y)dydu$.
