Page:Calculus Made Easy.pdf/217

Our instructions will then be:

${\begin{aligned}\int dy&=\int (x^{2}+x^{3})\,dx\\&=\int x^{2}\,dx+\int x^{3}\,dx\\y&={\tfrac {1}{3}}x^{3}+{\tfrac {1}{4}}x^{4}+C.\end{aligned}}$

If either of the terms had been a negative quantity, the corresponding term in the integral would have also been negative. So that differences are as readily dealt with as sums.

 

How to deal with Constant Terms.

 

Suppose there is in the expression to be integrated a constant term–such as this:

${\frac {dy}{dx}}=x^{n}+b$.

This is laughably easy. For you have only to remember that when you differentiated the expression $y=ax$, the result was ${\dfrac {dy}{dx}}=a$. Hence, when you work the other way and integrate, the constant reappears multiplied by $x$. So we get

${\begin{aligned}dy&=x^{n}\,dx+b\cdot dx,\\\int dy&=\int x^{n}\,dx+\int b\,dx,\\y&={\frac {1}{n+1}}x^{n+1}+bx+C.\end{aligned}}$

Here are a lot of examples on which to try your newly acquired powers.

 

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