MHB Compute $\int_{0}^{t}|s-t|\,ds$: Step-by-Step Guide

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To compute the integral $\int_{0}^{t}|s-t|\,ds$ for $t$ in the interval [0,1], note that on this interval, $s$ is always less than $t$, making $|s-t| = -(s-t)$. This simplifies the integral to $-\int_{0}^{t} (t-s) \, ds$. Evaluating this gives $\left[ts - \frac{s^2}{2}\right]_{0}^{t}$, resulting in the expression $t^2/2$. Thus, the final result of the integral is $\frac{t^2}{2}$.
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How would I compute the following integral? Let t be in [0,1].

$\int_{0}^{t}|s-t|\,ds$
 
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Fermat said:
How would I compute the following integral? Let t be in [0,1].

$\int_{0}^{t}|s-t|\,ds$

Since you are integrating on the interval $[0,t]$ so we have $0\leq s \leq t$.
 
Fermat said:
How would I compute the following integral? Let t be in [0,1].

$\int_{0}^{t}|s-t|\,ds$
The integral is for s from 0 to t so s is always less than t and s- t is always negative. |s- t|= -(s- t) so
\int_0^{t}|s- t| ds= -\int_0^t (t- s) ds= \left[ts- s^2/2\right]_0^t= t^2/2
 

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