Name: Rain Xia Student Number: 59118159
We can separate the area under curve into rectangles which have same width, f(t*)=t^2 is the height of each subinterval, we can get the area of each rectangles and sum them up to get the total area under the curve according to the definition of integral. In this question, when t=1,2,3, f(t)=0, there is no area in these cases, therefore we can just ignore them and count their area as 0. The total area does not affect by these bad points. Even if there is many removable discontinuities, the function is still integrable, we can receive the area under the curve by adding all the rectangles’ area together. The number of discontinuity point is limited, as n grows to infinity, the exist part of sum is also infinite. When we take the limit when n goes to 0, no matter how many discontinuities the function has, as long as it is infinite, the function is always integrable.