1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Thomas Nov 29 '15 at 19:50
  • $\begingroup$ @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. $\endgroup$ – user5539357 Nov 29 '15 at 19:55
  • $\begingroup$ no, it only says so if $F(a)=0$. $\endgroup$ – Thomas Nov 29 '15 at 19:56
  • $\begingroup$ Ok, my bad, I see. $\endgroup$ – user5539357 Nov 29 '15 at 19:58
  • $\begingroup$ 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. $\endgroup$ – user5539357 Nov 29 '15 at 20:00
1
$\begingroup$

You are probably referring to this part of the fundemental theorem of calculus \begin{equation} \frac{d}{dx} \int_a^x f(t)dt =f(x) \end{equation} And your using the fact that \begin{equation} f(x) = \frac{d}{dx} \int f(x) dx \end{equation} which yields \begin{equation} \frac{d}{dx} \int_a^x f(t)dt = \frac{d}{dx}\int f(x) dx \end{equation} 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.

$\endgroup$
1
$\begingroup$

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 $C \in \mathbb R$.

$\endgroup$
  • $\begingroup$ Dear @Thomas, you are correct indeed! Thanks for the suggestion, greatly improves the conceptual aspect too. $\endgroup$ – Lonidard Nov 29 '15 at 19:56
0
$\begingroup$

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)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.