Integration - the Australian Mathematical Sciences Institute

{16} • IntegrationIn fact, A ′ (x) = f (x). We can see this geometrically. To differentiate A(x) from first principles,we considerA(x + h) − A(x).hNow A(x +h) is the area under the graph up to x +h, and A(x) is the area under the graphup to x. Thus A(x + h) − A(x) is the area under the graph between x and x + h.yy = f(x)A(x+h)–A(x)0xx+hxIf you were to ‘level off’ the area under the graph between x and x + h, you’d obtain arectangle (as shown in the graph above) with width h and height equal to the averagevalue of f ; dividing the area by h gives the average value:A(x + h) − A(x)=harea of rectanglewidth of rectangle= height of rectangle= average value of f over the interval [x, x + h].If h becomes very small, then the interval [x, x + h] approaches the single point x. Andthe average value of f over this interval must approach f (x). So we haveA(x + h) − A(x)lim= f (x).h→0 hIn other words,A ′ (x) = f (x),and A(x) is an antiderivative of f (x).Returning again to our example f (x) = x 2 + 1, what then is the area function A(x)? Itsderivative must be f (x) = x 2 + 1. We’re looking for a function A such thatA ′ (x) = x 2 + 1 and A(0) = 0.