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I need to show this attribute for any $ x \in \mathbb{R}$ for regular integrals:

$$\int_0^x f(t)(x-t) \, dt = \int_0^x \left( \int_0^u f(t) \,dt \right) \,du$$

Well,my plan was to solve each of the sites..and show that they are equal.But that doesnt work. Have you any ideas?

Thank you :)

Michael Hardy
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wondering1123
  • 1,079

2 Answers2

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Define $F_1(x)$ and $F_2(x)$ as follows. $$F_1(x)=\int_{0}^{x}f(t)dt \space \text{ and }F_2(x)=\int_0^xF_1(t)dt$$ Now we can write RHS as $$\int_{0}^{x} F_1(u) - F_1(0)du = F_2(x)-F_2(0) -xF_1(x)$$ Now we can use Product rule to rewrite the LHS as follows $$x\int_0^x f(t)dt -\int_0^xtf(t)dt=\bigg(x(F_1(x)-F_1(0))\bigg) -\bigg(tF_1(t)\bigg|_0^x -\int_0^xF_1(t)dt\bigg)$$ $$=F_2(x)-F_2(0)-xF_1(x)$$ Same as we got on the RHS, and therefore we are done.

fractal1729
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Isaac Browne
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\begin{align} & \int_0^x \left( \int_0^u f(t) \,dt \right) \,du \\[10pt] = {} & \iint\limits_{\{ (t,u)\,:\, 0 \,\le\,t\,\le\,u\,\le\,x \}} f(t)\, d(t,u) \\[10pt] = {} & \int_0^x \left( \int_t^x f(t) \,du \right) dt \\[10pt] = {} & \int_0^x\left( f(t) \int_t^x 1\,du \right) dt & & \left(\begin{array}{c} \text{This can be done because $f(t)$ does} \\ \text{not change as $u$ goes from $t$ to $x$.} \end{array}\right) \\[10pt] = {} & \int_0^x f(t) (x-t) \, dt. \end{align}

Michael Hardy
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