Proof:
Define$$F(x) := \int_0^x f(t)\,dt.$$Since $f$ is continuous, the Fundamental Theorem of Calculus (FTC) tells us:$$F'(x) = f(x), \quad \text{for all } x \in [0,1].$$
Now define$$G(x) := \int_{1 - x}^1 f(1 - t)\,dt.$$We can rewrite this as:$$G(x) = \int_0^1 f(1 - t)\,dt - \int_0^{1 - x} f(1 - t)\,dt.$$
Differentiating both sides using the FTC and the chain rule, we get:$$G'(x) = -f(1 - (1 - x)) \cdot (-1) = f(x).$$
So $F'(x) = G'(x)$ for all $x \in [0,1]$, which implies that $F(x) - G(x)$ is constant.
Since $F(0) = 0$ and $G(0) = 0$, the constant is zero. Hence, $F(x) = G(x)$ for all $x$.
In particular, at $x = 1$,$$\int_0^1 f(t)\,dt = F(1) = G(1) = \int_0^1 f(1 - t)\,dt.$$
I don’t understand this proof at all, and I’d really appreciate some help.
Why do they define the two functions$$F(x) = \int_0^x f(t)\,dt \quad \text{and} \quad G(x) = \int_{1 - x}^1 f(1 - t)\,dt?$$
What’s the intuition behind introducing those, and how does that help prove the identity$$\int_0^1 f(t)\,dt = \int_0^1 f(1 - t)\,dt?$$
Also, why is this proposition even true? It feels very abstract to me. Is there a geometric or intuitive explanation of what this identity is really saying?
Finally, I’m having trouble understanding how the chain rule is used when differentiating $G(x)$. Could someone walk me through that part step-by-step?
Thank you!