Page:Calculus Made Easy.pdf/215

That is to say, $x^{2}dx$ will be changed ^[1] to ${\tfrac {1}{3}}x^{3}$. Put this into the equation; but don't forget to add the “constant of integration” $C$ at the end. So we get:

$y={\tfrac {1}{3}}x^{3}+C$.

You have actually performed the integration. How easy!

Let us try another simple case.

${\begin{aligned}Let\quad \quad \quad \quad {\dfrac {dy}{dx}}&=ax^{12},\end{aligned}}$

where $a$ is any constant multiplier. Well, we found when differentiating (see p. 29) that any constant factor in the value of $y$ reappeared unchanged in the value of ${\dfrac {dy}{dx}}$. In the reversed process of integrating, it will therefore also reappear in the value of $y$. So we may go to work as before, thus:

${\begin{aligned}dy&=ax^{12}\cdot dx,\\\int dy&=\int ax^{12}\cdot dx,\\\int dy&=a\int x^{12}\,dx,\\y&=a\times {\tfrac {1}{13}}x^{13}+C.\end{aligned}}$

So that is done. How easy!

1. You may ask: what has become of the little $dx$ at the end? Well, remember that it was really part of the differential coefficient, and when changed over to the right-hand side, as in the $x^{2}dx$, serves as a reminder that $x$ is the independent variable with respect to which the operation is to be effected; and, as the result of the product being totalled up, the power of $x$ has increased by one. You will soon become familiar with all this.