Mein Vorschlag:
\( \int_0^2 \, \frac y6 \, dy = \frac1{12} [x^2]_0^2=\frac4{12}=\frac13 \)
\( \frac1{12}\int_2^6\, 6-y \,dy = \frac1{12}[\,6y-\frac{y^2}{2}\,]_2^6 = \frac1{12}[\,6\cdot 6-\frac{6^2}{2}\,] -\frac1{12}[\,6\cdot 2-\frac{2^2}{2}\,]= \frac1{12}[\,36-18\,] -\frac1{12}[\,12-2\,] = \frac{18}{12}-\frac{10}{12}=\frac 8{12}= \frac23 \)
\( \int_0^2 \, \frac y6 \, dy + \frac1{12}\int_2^6\, 6-y \,dy =\frac 13 + \frac23=1 \)