# calculus – Calculating the integral with an indicator function

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) = intlimits_{max{0, ~t-1}}^{min{1,~ t}}x cdot mathbf 1_{0 le t-x le 1}~mathrm dx = begin{cases}displaystyleint_0^t x ~dx = frac{t^2}{2}, & t le 1 \(2mm) displaystyleint_{t-1}^1 x ~ dx = frac{2t – t^2}{2}, & t > 1end{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 problem comes from a probability problem