To show that a function with finitely many removable discontinuities is integrable, there is some basic knowledge we need to know: 1. a continuous function is always integrable on a close interval. 2. By doing an integral, we actually use lots of small rectangles to partly represent the actual area of a function, and finally take the limit of the width of these rectangles to get an accurate area. 3. A point has an area of zero.

Now when we draws many enough rectangles to represent the area of the function, since there are only finitely many x values at which the function is discontinuous, the rectangles they can “occupy” (means they are in that rectangles) are always finite, and will not increase even we change the rectangles we draw (Again, this is based on the fact we draw sufficiently many rectangles). Now when we begins to reduce the width of the rectangle, notice the number of rectangles occupied is always the same ,and thus the area of them is decreasing. When the width of a rectangle becomes zero by using limit, the whole curve is included and thus integrable except the places where the function is not continuous. However, at this position, the width of the rectangle has already been 0 ,and therefore, becomes a point. And a point has an area of zero ,which is not effected the original area of the function.