first, I must apologize for somewhat misleading a title.
To save both your and my time, I will go straight to the point.
By definition, an indefinite integral, or a primitive, or an antiderivative of a (some condition) function $f(x)$ is any $F(x)$ such that $F'(x)=f(x)$. All well and good.
Because any other primitive can be written as $F(x)+C$ for some constant $C$ (and this requires a proof), if we were to denote by $\int f(x)dx$ an antiderivative of $f(x)$, then
\begin{equation} \int f(x)dx = F(x)+C. \end{equation}
Fine. But here is the part that every textbook seems to have no problem with, but bugs me greatly: Often they say that integrate both sides of the following equation:
\begin{equation} f(x)=g(x),\end{equation}
to obtain
\begin{equation} \int f(x)dx = \int g(x)dx. \end{equation}
This looks like an ABSOLUTE nonsense to be for the following reason: IF both sides of the previous equation are TRULLY equal, then surely
\begin{equation} \int f(x)dx - \int g(x)dx =0. \end{equation}
But
\begin{equation} \int f(x)dx - \int g(x)dx =\int (f(x)-g(x))dx = \int 0dx, \end{equation}
which then equals $C$, any constant. Surely this is not necessarily 0!
So in short, this is my question: IS IT, STRICTLY SPEAKING, LEGAL, TO TAKE THE ANTIDERIVATIVE OF BOTH SIDES OF AN EQUATION?
Answer
The problem here might be the notation $\int f(x)\,dx$. Does it mean one primitive? All primitives? Something else?
If we for a while agree that $F$ is a primitive of $f$ on an interval $I$ and $G$ is a primitive of $g$ on the same interval $I$, and it holds that $f(x)=g(x)$ for all $x\in I$, then we can be sure that $F(x)=G(x)+C$ for all $x\in I$, where $C$ is some constant.
No comments:
Post a Comment