Question about derivative of an integral

[Mr Davis 97]

What does it mean to say that $\displaystyle\frac{d }{d x}\int f(x)dx = f(x)$? Does this somehow relate to the fundamental theorem of calculus? If so, how?

 

[Raghav Gupta]

What is the fundamental theorem according to you?

 

[jtbell]

What does it mean to say that $\displaystyle\frac{d }{d x}\int f(x)dx = f(x)$?

It means that if you find $$F(x) = \int {f(x)dx}$$ (i.e. the antiderivative, a.k.a. indefinite integral) and then you find $$\frac{d}{dx}F(x)$$ you get $f(x)$ back again.

 

[Mark44]

Another way to look at it is that differentiation and antidifferentiation are inverse operations.

 

[HallsofIvy]

The "Fundamental Theorem of Calculus" has two parts:
1) If we define $F(x)= \int_a^x f(x)dx$ then $dF/dx= f(x)$.
2) If f(x)= dF/dx then $F(x)= \int f(x) dx+$ some constant.

 

[Mr Davis 97]

The "Fundamental Theorem of Calculus" has two parts:
1) If we define $F(x)= \int_a^x f(x)dx$ then $dF/dx= f(x)$.
2) If f(x)= dF/dx then $F(x)= \int f(x) dx+$ some constant.

How are those two parts different? Also, isn't there a part that relates the definite integral to the antiderivative?

 

[SteamKing]

How are those two parts different? Also, isn't there a part that relates the definite integral to the antiderivative?

The "Fundamental Theorem of Calculus" has two parts:
1) If we define $F(x)= \int_a^x f(x)dx$ then $dF/dx= f(x)$.
2) If f(x)= dF/dx then $F(x)= \int f(x) dx+$ some constant.

In 1) above, F(x) is the antiderivative of f(x).

The First Fundamental Theorem of Calculus is more often expressed as:

$\int_a^b f(x)dx = F(b) - F(a)$, where F(x) is the antiderivative of f(x), as defined in part 2) above.

 

[WWGD]

Actually, the FTC says that $\frac {d}{dx} \int_0^x f(t)dt =f(x)$