Examining $\int f(x)dx=c$ $ \quad , c \in \mathbb{R}$.

Question

I am thinking of primative functions and this came to my mind. What if a indefinite integral equals to constant? Can i find a primative function? So i writed and tried to answer my own question. So could you help me if anywhere i have mistaken.Is my answer is true ? thanks in advance

Examine $\int f(x)dx=c$ $ \quad , c \in \mathbb{R}$.

I started with the sutiations

Case 1: if $\quad$ $c=0$ $\quad $ then

$\int f(x)dx=0 \iff f(x)=0$ as a result of linearity.

Case 2: if $\quad$ $c \neq 0 $ $\quad$ then

$\int f(x)dx=c \quad$ as a the result of fundemantal theorem of calculus we can rewrite

$\int f(x)dx=c = F(x)= \int_a^x f(x)dx$ $\quad$can be written. By picking the second and fourth part of the equility we can obtain

$c = \int_a^x f(x)dx$. Taking $\frac{d}{dx}$ of both sides and we obtain following,

$\frac{d}{dx} c=\frac{d}{dx}\int_a^x f(x)dx$ $\quad$ $\Rightarrow$ $\quad$ $0=f(x)$ which makes contradiction with our thesis.

By the result of this two cases if an indefinite integral is equals to a constant , the constant must be $0$.

Answer 1

Note that the primitive $F$ of a function $f$ is by definition a function whose derivative is the function. We denote by $\int f$ (or $\int f(x) dx$, no big deal here and $x$ is redundant so I omit it). Differentiation tells us that such a function is infinitely many due to the fact that $F +$ any constant is "a" $\int f$; that is why some author prefers to view $\int f$ as a set. With regard to calculation this difference does not matter. You can think of $\int f$ as a shorthand. Now again by definition $D\int f = f$; i.e. the derivative of the primitive of $f$ is $f$. So to answer your question there is no fundamental theorem of calculus involved. It involves only the definition of primitive.

Since we have $ D\int f = f $ by definition and $\int f = c$ by assumption and $Dc = 0$ by differentiation of a constant function, we have $$ f = 0. $$

The fundamental theorem of calculus is in essence to give the existence of a primitive under suitable conditions. Given a "nice" function $f$, how to give its primitive? The fundamental theorem of calculus says that the function $x \mapsto \int_{a}^{x}f$ is a handy choice.

Answer 2

$\int f(x)dx$ is an anti-derivative of $f$. Let $F(x)=\int f(x)dx$, then $F'(x)=f(x)$. If $F$ is constant, then $f(x)=0$ for all $x$ in the domain of $f$.