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

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SUMMARY

The integral $\int_{0}^{t}|s-t|\,ds$ for $t$ in the interval [0,1] evaluates to $\frac{t^2}{2}$. Since the integration occurs over the interval $[0,t]$, the expression $|s-t|$ simplifies to $-(s-t)$ because $s$ is always less than $t$. Thus, the integral can be rewritten as $-\int_0^t (t-s) ds$, leading to the final result after evaluation.

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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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