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192

Calculus Made Easy

until ascertained in some other way. So, if differentiating $x^{n}$ yields $nx^{n-1}$ , going backwards from ${\dfrac {dy}{dx}}=nx^{n-1}$ will give us $y=xn+C$ ; where $C$ stands for the yet undetermined possible constant.

Clearly, in dealing with powers of $x$ , the rule for working backwards will be: Increase the power by $1$ , then divide by that increased power, and add the undetermined constant.

So, in the case where

${\dfrac {dy}{dx}}=x^{n}$ ,

working backwards, we get

$y={\frac {1}{n+1}}x^{n+1}+C$ .

If differentiating the equation $y=ax^{n}$ gives us

${\frac {dy}{dx}}=anx^{n-1}$ ,

it is a matter of common sense that beginning with

${\frac {dy}{dx}}=anx^{n-1}$ ,

and reversing the process, will give us

$y=ax^{n}$ .

So, when we are dealing with a multiplying constant, we must simply put the constant as a multiplier of the result of the integration.

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