I am trying to figure out why we can rewrite
$\int_0^n s (\int_0^s 1 \, dt) \, ds = \frac{n^3}{3}$
as
$\int_0^n 1 (\int_s^n t \, dt) \, ds = \frac{n^3}{3}$
I would appreciate any pushes in the right direction. I'm not quite sure where to start.
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With the first integral do you mean $\int_0^n (\int_0^s s , ds), dt$? – A.P. Mar 26 '15 at 23:07
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No, I looked over both integrals and they are both what I meant to write. I added parenthesis now. – jason.tran Mar 26 '15 at 23:35
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Consider the square with vertices $(0,0),(0,n),(n,0),(n,n)$ in the $st$-plane and let $d$ be the diagonal from $(0,0)$ to $(n,n)$.
In the first integral you are summing the lengths of the vertical segments from the $s$-axis to $d$ multiplied by their abscissa.
On the other hand, in the second integral you are summing the lengths of the horizontal segments from the $t$-axis to $d$ multiplied by their ordinate.
Do you see that these operations are a mirror copy of each other?
A.P.
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