Prove: $\int_{0}^{x} f(u)(x-u) du = \int_{0}^{x} \left(\int_{0}^{u} f(t)dt\right)du$

Question

Prove: $$\int_{0}^{x} f(u)(x-u) du = \int_{0}^{x} \left(\int_{0}^{u} f(t)dt\right)du$$

I tried to differentiate both sides but then I get this

$$f(x)(x-x) = \int_{0}^{x} f(t)dt$$

Any help?

Answer 1

You didn't differentiate correctly - product rule is required.

$$\require{cancel}\int_{0}^{x} f(u)(x-u) du = \int_{0}^{x} \left(\int_{0}^{u} f(t)dt\right)du$$ Differentiating, $$\implies LHS=\frac{d}{dx}\left\{\int_0^xxf(u)\,du-\int_0^xuf(u)\,du\right\}=\int_0^xf(u)\,du+\cancel{xf(x)}-\cancel{xf(x)}\\\text{and }RHS=\frac{d}{dx}\left\{\int_{0}^{x} \left(\int_{0}^{u} f(t)dt\right)du\right\}=\int_0^xf(t)\,dt$$

So they have the same derivative, and at $x=0$, both are the zero, hence done.

Answer 2

If you apply Fubini's theorem to the right-hand side, the result is $$ \int_0^x\Big(\int_0^uf(t)\;dt\Big)du=\int_0^x\Big(\int_t^x\;du\Big)f(t)\;dt=\int_0^x(x-t)f(t)\;dt$$

Alternately, you could use Leibniz's rule to differentiate both sides, noting that both sides are zero when $x=0$.

Answer 3

Hint: Use Fubini's theorem to conclude that $$\int_{0}^{x} \int_{0}^{u} f(t)\,\text{d}t\,\text{d}u = \int_{0}^{x} \int_{u}^{x} f(t)\,\text{d}u\,\text{d}t,$$ from where the result follows.