I understand that as a consequence of the fundamental theorem of calculus it is true that

$$\frac{d}{dx}\int_a^x f(t)dt=f(x)$$

Reading an intro ODE text the very first chapter says that

$$\frac{dy}{dx}=f(x)\implies y(x)=\int f(x)dx+C$$

by integrating on both sides. I assume this is a consequence of the FTC, but I don't quiet see it. If I integrate on both sides I get

$$\frac{dy}{dx}=f(x)\implies \int \frac{d}{dx}y(x)dx=\int f(x)dx$$

at this point to apply the FTC I have to make an assumption about the interchangeability of the integral and derivative operators, which hopefully only requires $\frac{d}{dx}y(x)$ to be integrable, I am not sure but willing to overlook that part for now. So I have

$$ \frac{d}{dx} \int y(x)dx=\int f(x)dx$$

But to be able to actually claim that the integral and the derivative cancel, I need to make the integral definite:

$$ y(x)=\frac{d}{dx} \int_a^x y(t)dt=\int_a^x f(x)dx$$

At this point for the text to make sense, I'm guessing it is implicitly assumed when I see indefinite integrals in PDE/ODE textbooks that all statements about them are meant to hold true for any arbitrary interval in the domain, which can be represented by $[a,x]$. Is that reasonable? Am I missing something obvious? Thanks

  • $\begingroup$ Your very first equation says that $y(x)=\int_a^xf$ is a solution of the DE. $\endgroup$ – symplectomorphic Jun 16 '17 at 5:23
  • $\begingroup$ And as for your question about needing to interchange differentiation and integration: all that is beside the point. On the left hand side of the consequent of your third line, you just perform a change of variables; setting $u=y$ gives $du=y'(x)dx$. $\endgroup$ – symplectomorphic Jun 16 '17 at 5:28

By definition, the indefinite integral symbol $\int f(x) \, dx$ denotes the most general function (or the set of all functions) whose derivative is $f(x)$, so the implication in question is a completely trivial matter of notation; it has nothing at all to do with the FTC.

(And the “$+C$” is superfluous, since the constant is implicit in the notation $\int f(x) \, dx$.)

  • $\begingroup$ This is the answer which is simple as well as correct. The meaning of symbol for indefinite integration is exactly what is given here. +1 $\endgroup$ – Paramanand Singh Jun 16 '17 at 13:52

Let $F$ be a primitive function of $f$, i.e. $\frac d{dx}F = f$. By the fundamental theorem (Newton-Leibniz formula), we have $$\int_a^xf(t)\,dt = F(x) - F(a).$$

In your case, $y$ is primitive function of $f$, and thus $$\int_a^xf(t)\,dt = y(x) - y(a)\implies y(x) = \int f(x)\,dx+C.$$

  • $\begingroup$ What happend with $y(a)$ in your last implication? Is $y(a)=C$? $\endgroup$ – JDoeDoe Jul 16 '17 at 7:59
  • $\begingroup$ I just switched notation to match OP's, @JDoeDoe. $\endgroup$ – Ennar Jul 16 '17 at 8:34

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