Page:Calculus Made Easy.pdf/216

We begin to realize now that integrating is a process of finding our way back, as compared with differentiating. If ever, during differentiating, we have found any particular expression–in this example $ax^{12}$–we can find our way back to the $y$ from which it was derived. The contrast between the two processes may be illustrated by the following remark due to a well-known teacher. If a stranger were set down in Trafalgar Square, and told to find his way to Euston Station, he might find the task hopeless. But if he had previously been personally conducted from Euston Station to Trafalgar Square, it would be comparatively easy to him to find his way back to Euston Station.

 

Integration of the Sum or Difference of two Functions.

 

${\begin{aligned}{\mbox{Let }}\quad \quad \quad \quad {\frac {dy}{dx}}&=x^{2}+x^{3},\\{\mbox{then }}\quad \quad \quad \quad dy&=x^{2}\,dx+x^{3}\,dx.\end{aligned}}$

There is no reason why we should not integrate each term separately: for, as may be seen on p. 35, we found that when we differentiated the sum of two separate functions, the differential coefficient was simply the sum of the two separate differentiations. So, when we work backwards, integrating, the integration will be simply the sum of the two separate integrations.