$f:[0,1] \rightarrow \mathbb{R}$ 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.


My attempt.
I evaluated by the calculator that $\int_{0}^{1} \int_{x}^{1-x} f(y) d y d x=0.$
When we make the variable change $u=1-x, x=0, u=1, x=1, u=0, d u=-d x \int_{0}^{1} \int_{x}^{1-x} f(y) d y d x$
$=\int_{1}^{0} \int_{1-u}^{u} f(y) d y(-d u)=\int_{1}^{0} \int_{u}^{1-u} f(y) d y d u=-\int_{0}^{1} \int_{u}^{1-u} f(y) d y d u$
Then, I couln't continue, can you help? Thanks...
 A: Since $f$ is continuous, it has an antiderivative $F: [0,1] \to \mathbb R$, and so $\int_x^{1-x} F(y) dy = F(1-x)-F(x)$. So your integral is equal to
$$\int_0^1 (F(1-x)-F(x))dx = \int_0^1 F(1-x) dx - \int_0^1 F(x) dx
$$
Using a change of variables $u=1-x$ in the first term, and for kicks using a change of variables $u=x$ in the second term, you'll see that you get $0$.
A: Let $I$ be your original integral. You have shown that $I=-I$. But this implies that $2I=0$ i.e. $I=0$. 
Another way of proving the result is splitting the integral at $1/2$.
A: Note: There's something very suspicious about this problem. Note that $x\le 1-x$ on the interval $[0,1/2]$, whereas on the interval $[1,2/1]$, when we write $\int_x^{1-x}\,dx$ we really need to reverse the limits of integration and introduce a negative sign.
Since $f$ is continuous, we can apply Fubini's Theorem and interchange the order of integration. Note that each integral needs to be split into two.
\begin{align*}
\int_0^1\int_x^{1-x} f(y)\,dy\,dx &= \int_0^{1/2}\int_x^{1-x} f(y)\,dy\,dx - \int_{1/2}^1\int_{1-x}^x f(y)\,dy\,dx \\ 
&=\int_0^{1/2}\int_0^y f(y)\,dx\,dy + \int_{1/2}^1\int_0^{1-y} f(y)\,dx\,dy \\ &\hspace{.5in}-\int_0^{1/2}\int_{1-y}^1 f(y)\,dx\,dy - \int_{1/2}^1\int_y^1 f(y)\,dx\,dy \\
&=\int_0^{1/2} yf(y)\,dy + \int_{1/2}^1(1-y)f(y)\,dy \\
&\hspace{.5in} - \int_0^{1/2}(1-(1-y))f(y)\,dy - \int_{1/2}^1 (1-y)f(y)\,dy \\
&=\int_0^{1/2} \big(f(y)-f(y)\big)\,dy + \int_{1/2}^1 \big((1-y)-(1-y)\big)f(y)\,dy  \\ &= 0.
\end{align*}
Eureka!
