Triple integral problem involving definite integrals (and Taylor's formula possibly) Any hint on how to approach this question?
Show that $$\int_0^x\int_0^v\int_0^u f(t)dtdudv=\frac12\int_0^x(x-t)^2f(t)dt$$
I am completely clueless.
I tried to convert it into the standard xyz form so I get
$$\int_0^a\int_0^x\int_0^y f(z)dzdydx=\frac12\int_0^a(a-z)^2f(z)dz$$
But it doesn't help much...
Update: well, and I change the order of integration of the inside double integral and got $$\int_0^a\int_0^x\int_z^x f(y)dydzdx$$, is this a right move?
 A: Chaning the order of integration is the right move, but you need to do it in a way that makes it possible to move the function $f$ out twice. Our desired order of integration is thus $$
  \iiint  f(t) \,du\,dv\, dt \text{ or } \iiint  f(t) \,dv\,du\, dt
$$
Now we just need to find the new integration limits. Your original integral integrates over the set $$
  A_x = \left\{(t,u,v) \,:\, v \in [0,x], u \in [0,v], t \in [0,u] \right\}
$$
meaning $$
  A_x = \left\{(t,u,v) \,:\, 0 \leq t \leq u \leq v \leq x \right\} \text{.}
$$
Thus, the reordered integral, including limits, is $$\begin{eqnarray}
  \int_0^x \int_t^x \int_u^x f(t) \,dv\,du\,dt
  &=& \int_0^x f(t) \int_t^x \int_u^x 1 \,dv\,du\,dt \\
  &=& \int_0^x f(t) \int_t^x (x-u)\,du\,dt \\
  &=& \int_0^x f(t) \left(xu-\tfrac{1}{2}u^2\right)\big|_{u=t}^{u=x} \,dt \\
  &=& \int_0^x f(t) \underbrace{\left(x^2 - \tfrac{1}{2}x^2 -xt+\tfrac{1}{2}t^2\right)}_{=\frac{1}{2}(x^2-2xt + t^2) = \frac{1}{2}(x - t)^2} \,dt \\
  &=& \frac{1}{2}\int_0^x f(t) (x - t)^2 \,dt \text{.}
\end{eqnarray}$$
A: $$\int_0^x\int_0^v\int_0^u f(t)dtdudv$$after changing order of integration comes to be as:$$\int_0^x\int_t^x\int_u^x f(t)dvdudt$$
on solving:
u get $$\frac12\int_0^x(x-t)^2f(t)dt$$
