For a function whose graph is always above the x-axis, the integral can be thought of as "area under the curve". The thing to be careful about with this problem is that part of the graph is below the x-axis. That area has to be thought of as "negative area".Question 2
Since each of the vertical lines represents a step of 1/2, from x=0 to x= 1, the graph is below the x-axis. The area is that of a right triangle with base 1 and height 1. That area will be negative (if you like, think of it as a triangle with base 1 and height -1).
For x from 1 to 6, the graph is above the x-axis and the area is that of a trapezoid with height 1 and bases of length 5 and 3.
For x from 6 to 7, the graph is again below the x-axis the area is exactly the same as the first triangle.
Use geometric area formulas to find those areas.