Difference between integral and a function

[kolleamm]

Online it says that the integral is the opposite of the derivative. So x^2 is the integral of 2x.

So if f(x) = x^2 , does that mean that the integral is just the function itself? Basically whatever f(x) equals?

Thanks in advance

 

[phinds]

Online it says that the integral is the opposite of the derivative. So x^2 is the integral of 2x.

No, the integral of 2x is x^2 + c. They are not exactly symmetric.

So if f(x) = x^2 , does that mean that the integral is just the function itself? Basically whatever f(x) equals?

No, for the reason just stated.

 

[Drakkith]

Both x² and 2x are functions, and they are different functions. The two are related through integration and differentiation, as explained by the first fundamental theorem of calculus.

To be clear, we should label the two functions differently:

F(x) = x²
f(x) = 2x

F(x) and f(x) are different functions, with f(x) being the derivative of F(x). Stated in math terminology, F'(x) = f(x) and ∫f(x)dx = F(x).

 

[fresh_42]

Online it says that the integral is the opposite of the derivative. So x^2 is the integral of 2x.

So if f(x) = x^2 , does that mean that the integral is just the function itself? Basically whatever f(x) equals?

Thanks in advance

You're question is a bit confusing. So let's start with a smooth function $f: \mathbb{R} \rightarrow \mathbb{R}$, e.g. $f(x)=x^2$.
(I picked smoothness here for reason of simplicity and it means mathematically basically the same as in ordinary language: no gaps or corners, pretty smooth.)

Then the derivative $f\,'$ of $f$ is again a smooth function that associates to each point $x$ the slope of $f$ which is defined as the steepness of its tangent, i.e. the steepness of the touching line. In the example we have $f\,'(x)=2x$ which means, that e.g. at the point $x=5$ we have a slope of $2\cdot 5 = 10$ ($10$ inches of gained height on $1$ inch horizontal deviation).

The integral $\int f(x)\, dx = F(x)$ of $f$ is a function $F$ that measures the area between the $x-$axis and the function $f$. Basically as we measure the area of a room: length times width. Only that it also works for curved walls. In the example we have $\int x^2 dx = \frac{1}{3}x^3$ which means, that the area $A$ underneath the parabola $x^2$, say between the points $x=1$ and $x=4$, is $A= \frac{1}{3}\cdot 4^3 - \frac{1}{3} 1^3 = 21$.

Now what makes a derivative and an integral sort of "opposite" is, that $\int f\,'(x) dx = f(x)$, or in our example $\int 2x \,dx= x^2$.

 

[kolleamm]

thank you for your help although I'm still confused