# Antiderivative and definite integral

If $f$ is a continuous, real-valued function on interval $[a,b]$, then the fundamental theorem of calculus tells us that

$$\int_a^x f(t)dt=F(x)$$

where $F(x)$ is antiderivative, i.e. $F(x)'=f(x)$.

If so, why I can't find the equality $\int f(x)dx=\int_a^x f(t)dt=F(x)$ anywhere? It expresses the relationship between definite and indefinite integral in such a straightfoward way (assuming this equality is true). So it it true and can I use $\int$ and $\int_a^x$ interchangeably?

• Actually the theorem says (if $f$ is continuous, then) $$\int_a^x \,f(t)dt = F(x)-F(a)$$ which differs from what you have written down in two essential aspects. – Thomas Nov 29 '15 at 19:50
• @Thomas I actually mean the first part of the fundamental theorem of calculus (mixed up x with t, sorry for that) - it says that $\int_a^x f(t)dt = F(x)$., and it's different from what you have in your comment. – user5539357 Nov 29 '15 at 19:55
• no, it only says so if $F(a)=0$. – Thomas Nov 29 '15 at 19:56
• Ok, my bad, I see. – user5539357 Nov 29 '15 at 19:58
• So the point is I can't use them interchangeably, right? Because the first one gives one particular primitive function, and the second one means any. – user5539357 Nov 29 '15 at 20:00

You are probably referring to this part of the fundemental theorem of calculus $$\frac{d}{dx} \int_a^x f(t)dt =f(x)$$ And your using the fact that $$f(x) = \frac{d}{dx} \int f(x) dx$$ which yields $$\frac{d}{dx} \int_a^x f(t)dt = \frac{d}{dx}\int f(x) dx$$ While this is true, remember that just because two functions have the same derivative, doesn't mean that they're the same function. Here is an example \begin{align} f(x) &= \int_3^x \sin t dt = -\cos x + \cos 3\\ g(x) &= \int \sin x dx = -\cos x \end{align} As you can cleary see both functions have the same derivative, but the functions are not the same. Your statement will be true if $a$ is a root of the indefinite integral. Example \begin{align} \int_0^x 2t dt= x^2 - 0 = x^2 = \int 2x dx \end{align} So it really depends on the function you are using. But the statement is NOT always true.

Actually, the Fundamental Theorem of Calculus states that if $f:[a,b]\to \mathbb R$ is a continuous real-valued function, and $F$ is its antiderivative, then

$$\int_a^x f(t)dt= F(x)-F(a)$$

This is precisely why, more accurately,

$$\int_a^x f(t)dt= F(x)-F(a)=\int f(x)dx +C$$

for some $C \in \mathbb R$.

• Dear @Thomas, you are correct indeed! Thanks for the suggestion, greatly improves the conceptual aspect too. – Lonidard Nov 29 '15 at 19:56

For a F(x) fixed, it's only ONE primitive function of f.

the other functions are of the form F(x) + arbitrary constante.

The form $\int_a^x f(u)du$ is the primitive which is 0 at x=a.

So $\int f(u) du = F(x)+ C$ where $C$ is an arbitrary number.

( the derivative of C is 0)