Fundamental Theorem of Calculus confusion

Question

I know that $\frac{d}{dx} \int_{0}^{x} f(t) dt = f(x)$. What about $\frac{d}{dx} \int_{0}^{x} f(t-x) dt$? Is that just $f(x-x)$? I think I have to use the chain rule but I'm not sure how.

Thank you!

Answer 1

The following is an informal way of figuring out what the answer is. For someone like me who has a limited number of brain cells, it beats trying to remember a formula.

Let $F(u)$ be any antiderivative (indefinite integral) of $f(u)$. Then $$\int_0^x f(t-x)\,dt=\left.F(t-x)\right|_0^x=F(0)-F(-x).$$ Now differentiate with respect to $x$. By the definition of antiderivative, we have $F'(u)=f(u)$. Thus, by the Chain Rule, our derivative is $-(-f(-x))$, or more simply $f(-x)$.

Answer 2

Another approach would be the following:

$$\int\limits_0^x {f\left( {t - x} \right)dt} \overbrace = ^{t - x = u}\int\limits_{ - x}^0 {f\left( u \right)du} = - \int\limits_0^{ - x} {f\left( u \right)du} $$

From this it is immediate that

$$\frac d{dx}\int\limits_0^x {f\left( {x - t} \right)dt} =-\frac d{dx} \int\limits_0^{-x} {f\left( u \right)du}=f(-x) $$

where the $-$ signs cancelled.