Simply find the area under the velocity curve. Over the first 3 second interval, the area is that of a trapezoid with bases 2 and 5, so that area/distance traveled is [tex]\left(\dfrac{2+5}2\dfrac{\rm m}{\rm s}\right)\cdot(3\,\mathrm s)=\dfrac{21}2\,\mathrm m[/tex]. For the last 4 seconds, the area/distance is [tex]\left(5\dfrac{\rm m}{\rm s}\right)(4\,\mathrm s)=20\,\mathrm m[/tex].
Then the total distance traveled is
[tex]\dfrac{21}2\,\mathrm m+20\,\mathrm m=\dfrac{61}2\,\mathrm m=30.5\,\mathrm m[/tex]
