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Compute $\displaystyle\int\limits_{t-2T}^T t-\tau\cdot 1 \;d\tau$

\begin{align} \displaystyle&\int\limits_{t-2T}^T t-\tau\cdot 1 \;d\tau \\ &=\left[t\tau - \frac{\tau^2}{2}\right]_{t-2T}^{T}\\ &=tT-\frac{T^2}{2}-t(t-2T)-\frac12(t-2T)^2\\ &=tT-\frac{T^2}{2}-t^2+2tT-\frac 12(t^2-4tT+4T^2)\\ &=tT-\frac{T^2}{2}-t^2+2tT-\frac 1 2t^2+2Tt-2T^2\\ &=\frac{-3T^2}{2}+5tT-\frac{5tT^2}{2} \end{align}

Is this correct?

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    $\begingroup$ I got this here $$1/2\, \left( T+t \right) \left( 3\,T-t \right) $$ $\endgroup$ – Dr. Sonnhard Graubner May 4 at 10:31
  • $\begingroup$ @Dr.SonnhardGraubner can you share your process to the solution? $\endgroup$ – Doesbaddel May 4 at 10:37
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    $\begingroup$ In evaluating the antiderivative you made a plus-minus mistake. It should be $+(t-2T)^2/2$ instead of what you have. $\endgroup$ – Stan Tendijck May 4 at 10:40
  • $\begingroup$ Who uses $t$, $\tau$, and $T$ in the same expression? Yikes $\endgroup$ – Matthew Leingang May 4 at 10:40
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    $\begingroup$ Don't worry, I'm not blaming you. $\endgroup$ – Matthew Leingang May 4 at 10:46
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We have $$[t\tau-\tau^2/2]_{t-2T}^{T}=tT-\frac{T^2}{2}-\left(t(t-2T)-\frac{(t-2T)^2}{2}\right)$$ and this is $$tT-\frac{T^2}{2}-t^2+2tT+\frac{1}{2}t^2-2tT+2T^2$$

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  • $\begingroup$ Thank you for pointing out that mistake. $\endgroup$ – Doesbaddel May 4 at 10:46

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