Derivative of indefinite integral vs. definite

So I understand what to do if I have $$\frac{d}{dx} \int_a^b f(x)dx$$ for some function $$f$$.

But if I have $$\frac{d}{dx} \int f(x)dx$$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?

And if I have the $$\int \frac{d}{dx} f(x)dx$$ - it is just the function as well, correct?

Thanks

• You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus – Ethan Bolker Dec 16 '18 at 1:20
• "So I understand what to do if I have $\frac{d}{dx} \int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way. – user587192 Dec 16 '18 at 2:35

Not quite. The arbitrary constant from integration changes it up. Explicitly,

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

$$\int \left( \frac{d}{dx} f(x) \right) dx = f(x) + C$$

An example will be illustrative in finding the anomaly if it's not obvious. Let $$f(x) = x^2$$. We know $$\int f(x)dx = x^3/3 + C$$, $$f'(x) = 2x$$, and, obviously, the derivative of a constant $$C$$ is $$0$$.

Then

$$\frac{d}{dx} \int f(x)dx = \frac{d}{dx} \int x^2dx = \frac{d}{dx} \left( \frac{x^3}{3} + C \right) = \frac{3x^2}{3} + 0 = x^2 = f(x)$$

but

$$\int \left( \frac{d}{dx} f(x) \right) dx = \int \left( \frac{d}{dx} x^2 \right) dx = \int 2xdx = \frac{2x^2}{2} + C = x^2 + C = f(x) + C$$

For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.

For the second question, you have to add the Constant of Integration to the function because of the rules of integration.