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I'm having a problem calculating $$f(t) = \int_0^1 x \cdot \mathbf 1_{0 \le t-x \le 1}~\mathrm dx$$

The indicator inequality implies $t-1 \le x \le t$. Firstly, if $t \le 0$, then $t - x \le 0$ so $f(t) = 0$. Similarly, $f(t) = 0$ when $t > 2$. Now we are left with $t \in (0,2]$. We now have $$f(t) = \int\limits_{\max\{0, ~t-1\}}^{\min\{1,~ t\}}x \cdot \mathbf 1_{0 \le t-x \le 1}~\mathrm dx = \begin{cases}\displaystyle\int_0^t x ~\mathrm dx = \frac{t^2}{2}, & t \le 1 \\[2mm] \displaystyle\int_{t-1}^1 x ~ \mathrm dx = \frac{2t - t^2}{2}, & t > 1\end{cases}$$ Therefore, $$f(t) = \begin{cases}0, & t \in (-\infty, 0] \cup (2,+\infty) \\[2mm] \dfrac{t^2}{2}, & t \in (0,1] \\[2mm] \dfrac{2t-t^2}{2}, & t \in (1, 2]\end{cases}$$ Is this correct?

Note: I am adding the probability tag because this integral comes from a probability problem

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Your result is correct, but there is a bit of imprecision in your work, since you cannot write the limits of integration as simply $\max(0, t-1)$ and $\min(1, t)$. Rather, they would need to be written as $$\min(\max(0, t-1), 1), \quad \max(\min(1, t), 0).$$

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  • $\begingroup$ I think I got your point. Thanks for the clarification and verification. $\endgroup$ – NodeJS Jun 10 at 20:56

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