How is this function integrated?

I'm looking for some clarification on why each strip has area $(2T - |u|)du$ in the excerpt below, I'm not sure how the change of variables is done.

• It's just the area of a trapezium. Dec 5, 2017 at 3:07
• @LordSharktheUnknown I still don't see how they get their formula from the formula for the area of a trapezoid; could you please explicitly connect them? Dec 5, 2017 at 3:13

Let us deal with the case $"du>0"$. I guess it's just big triangle minus small triangle. Fix $u$. In the picture, the big triangle is the area underneath $u=t-t'$ which is \begin{align} \frac{1}{2}(2T-u)^2 \end{align} since when we set $t'=-T$ the corresponding $t$ value is $t = u-T$. Likewise, we see that the small triangle is the area underneath $u+du= t-t'$ which is \begin{align} \frac{1}{2}(2T-u-du)^2. \end{align} Thus, the area of the strip is \begin{align} \frac{1}{2}[(2T-u)^2-(2T-u-du)^2] = \frac{1}{2}(4T-2u-du)du = \frac{1}{2}(4T-2u)du + \text{lower order term}. \end{align}