Page:Calculus Made Easy.pdf/204

 we stop, there will still be a piece wanting to make up the whole $$2$$ inches; and the piece wanting will always be the same size as the last piece added. Thus, if after having put together $$1$$, $$\tfrac{1}{2}$$, and $$\tfrac{1}{4}$$, we stop, there will be $$\tfrac{1}{4}$$ wanting. If we go on till we have added $$\tfrac{1}{64}$$, there will still be $$\tfrac{1}{64}$$ wanting. The remainder needed will always be equal to the last term added. By an infinite number of operations only should we reach the actual $$2$$ inches. Practically we should reach it when we got to pieces so small that they could not be drawn–that would be after about $$10$$ terms, for the eleventh term is $$\tfrac{1}{1024}$$. If we want to go so far that not even a Whitworth’s measuring machine would detect it, we should merely have to go to about $$20$$ terms. A microscope would not show even the $$18^{th}$$ term! So the infinite number of operations is no such dreadful thing after all. The integral is simply the whole lot. But, as we shall see, there are cases in which the integral calculus enables us to get at the exact total that there would be as the result of an infinite number of operations. In such cases the integral calculus gives us a rapid and easy way of getting at a result that would otherwise require an interminable lot of elaborate working out. So we had best lose no time in learning how to integrate.