Page:Calculus Made Easy.pdf/218

Examples

(1) Given ${\displaystyle {\dfrac {dy}{dx}}=24x^{11}}$. Find ${\displaystyle y}$. Ans. ${\displaystyle y=x^{3}2+C}$.

(2) Find ${\displaystyle \int (a+b)(x+1)\,dx}$. It is ${\displaystyle (a+b)\int (x+1)\,dx}$ or ${\displaystyle (a+b)\left[\int x\,dx+\int dx\right]}$ or ${\displaystyle (a+b)\left({\dfrac {x^{2}}{2}}+x\right)+C}$.

(3) Given ${\displaystyle {\dfrac {du}{dt}}=gt^{\frac {1}{2}}}$. Find ${\displaystyle u}$. Ans. ${\displaystyle u={\frac {2}{3}}gt^{\frac {3}{2}}+C}$.

(4) ${\displaystyle {\dfrac {dy}{dx}}=x^{3}-x^{2}+x}$. Find ${\displaystyle y}$.

and ${\displaystyle \quad \quad \quad \quad y={\tfrac {1}{4}}x^{4}-{\tfrac {1}{3}}x^{3}+{\tfrac {1}{2}}x^{2}+C}$.

(5) Integrate ${\displaystyle 9.75x^{2.25}\,dx}$. Ans. ${\displaystyle y=3x^{3.25}+C}$.

All these are easy enough. Let us try another case.

Let ${\displaystyle \quad \quad \quad \quad {\dfrac {dy}{dx}}=ax^{-1}}$.

Proceeding as before, we will write

Well, but what is the integral of ${\displaystyle x^{-1}\,dx}$?

If you look back amongst the results of differentiating ${\displaystyle x^{2}}$ and ${\displaystyle x^{3}}$ and ${\displaystyle x^{n}}$, etc., you will find we never got ${\displaystyle x^{-1}}$ from any one of them as the value of ${\displaystyle {\dfrac {dy}{dx}}}$. We got ${\displaystyle 3x^{2}}$ from ${\displaystyle x^{3}}$; we got ${\displaystyle 2x}$ from ${\displaystyle x^{2}}$; we got ${\displaystyle 1}$ from ${\displaystyle x^{1}}$ (that is, from ${\displaystyle x}$ itself); but we did not get ${\displaystyle x^{-1}}$ from ${\displaystyle x^{0}}$, for two very good reasons. First, ${\displaystyle x^{0}}$ is simply =1, and is a constant, and could not have