Triple integral: $\int_0^{x} \int_{a-x}^{a} \int_0^{b} f(c ) \ dc \ db \ da$

I am given the following triple-integral and some parameter $x > 0$

$$\int_0^{x} \int_{a-x}^{a} \int_0^{b} f(c ) \ dc \ db \ da$$

and I am asking myself whether I can rewrite this integral in the following way:

It seems, that the variable $c$ can vary between $-x$ and $x$. Thus, I am asking myself whether we can rewrite this integral as

$$\int_{-x}^{x} g(c) f(c ) \ dc$$

where $g(c)$ is a function that is to be determined.

• Yes, by changing the order of integration so that $dc$ is on the outside. – kccu May 17 '18 at 21:34

Let $F(t) = \int_0^t f(u) du$ and $J$ your integral. Then $$J = \int_0^x \int_{a-x}^a F(b) db\ da.$$ The region of integration is made of two triangles, say $T_1=co\{(0,0),(x,0),(0,-x)\}$ and $T_2=co\{(0,0),(x,0),(x,x)\}$. By exchanging the integrals and then integrating by parts you get \begin{align*} \int_{T_1} F(b) da\ db &= \int_{-x}^0 (b+x)F(b) db = \left[ F(b)\left(\frac{b^2}{2}+bx\right)\right]_{-x}^0 - \int_{-x}^0 f(b)\left(\frac{b^2}{2}+bx\right) db\\ &= \frac{x^2}{2}F(-x) - \int_{-x}^0 f(b)\left(\frac{b^2}{2} +bx\right) db\\ &= -\frac{1}{2} \int_{-x}^0 f(b) (x^2+2bx+b^2) db \end{align*} and \begin{align*} \int_{T_2} F(b) da\ db &= \int_0^x (x-b)F(b) db = \left[ F(b)\left(bx-\frac{b^2}{2}\right)\right]_0^x - \int_0^x f(b)\left(bx-\frac{b^2}{2}\right) db\\ &= \frac{x^2}{2}F(x) - \int_0^x f(b)\left(bx-\frac{b^2}{2}\right) db\\ &= \frac{1}{2} \int_0^x f(b) (x^2-2bx+b^2) db \end{align*}
This shows that the desired function $g$ is $$g(c) = \begin{cases} -\frac{1}{2}(x+c)^2 & \text{if c < 0} \\ \frac{1}{2}(x-c)^2 & \text{if c \geq 0.} \end{cases}$$