Are derivatives and integrals inverses? \begin{align}f(x) &= x^2\\
g(x)&=\int_0^xt^2dt = \frac{x^3}{3}\\
\end{align} 
$(2,4)$ is a point on $f(x)$, therefore $(4,2)$ must exist on its inverse.
However $g(4) = 64/3\ne 2$.
Also clearly $x^2$ and $\frac{x^3}{3}$ are not inverses. What am I missing?
 A: You seem to mix the concepts of inverse functions and inverse operations:


*

*for the functions $\mathbb{R^+}\to\mathbb{R^+}$ given by $f(x)=x^2$ and $g(x)=\sqrt{x}$, you have:
$$\color{blue}{2} \xrightarrow{f} \color{red}{4} \xrightarrow{g} \color{blue}{2}$$
so in your wording: $(2,4)$ belongs to $f$ and $(4,2)$ belongs to $g$;

*for the operations "integrate" $\int$ (indefinite; i.e. get an antiderivative) and "differentiate" $\frac{d}{dx}$, you have:
$$\color{blue}{x^2} \xrightarrow{\displaystyle\int} \color{red}{\frac{x^3}{3}  \;(\,+\,C\,)}\xrightarrow{\displaystyle\frac{d}{dx}} \color{blue}{x^2}$$
so in your wording you could say that $(x^2,\frac{x^3}{3})$ belongs to (the operation) $\int$ and $(\frac{x^3}{3},x^2)$ belongs to (the operation) $\frac{d}{dx}$.
So it's similar in a way, but the functions act on numbers while the operations above act on functions.
A: As was stated in the comments, you're confusing integration and differentiation with inverse functions. An example of inverse functions would be $y=2x$ and $\frac{1}{2}x=y$. Inverse functions are, in a simple sense, reflections over the line $y=x$. Inverse operations, however, are operations that "undo" each other.  Integration and differentiation are inverse operations in the same way addition and subtraction are inverse operations (notice the difference between functions and operations). They are inverses because they "undo" each other. Let's take a look at addition and subtraction before moving up to integrals and derivatives. If I have a number, $n$, and I add five, I have the expression $5+n$. However, if I subtract five from that expression, I have $5+n-5$, which is equal to $n$. Notice that we are left with the original number.
Now, integration and differentiation are inverses in the same exact way. Take the expression $x^2$. If I differentiate $x^2$, I get $2x$. However, if I integrate $2x$, I get $x^2$. Notice that we are left with the original expression. Integration and differentiation "undo" each other.
A: You're confusing two terms together: inverse operations and inverse functions. They are different concepts. Differentiation and integration are inverse operations just like addition and subtraction are. They're not inverse functions.
Inverse functions are functions that satisfy the following condition for all $x$ that are in the domains of the functions (an inverse function is typically denoted by $f^{-1}(x)$):
$$f^{-1}(f(x))=x\ \text{or}\ f(f^{-1}(x))=x.$$
An important thing to note that graphically inverse functions are symmetric with respect to the line $y=x$.
For example, the functions $f(x)=2^x$ and $f^{-1}(x)=\log_2{x}$ are inverse operations because:
$$
\log_2{2^x}=x\ (x\in R)\ \text{and}\ 2^{\log_2{x}}=x\ (x>0).
$$
If you graph these two functions, you will see that their graphs are symmetric with respect to $y=x$.
Inverse operations, on the other hand, simply undo each other (in street language, one of the operations is the backwards version of the other):
$$
\frac{d}{dx}\left(\int x^2\,dx\right)=\frac{d}{dx}\left(\frac{x^3}{3}+C\right)=x^2.
$$
To put it simply, you get back what you started with.
If you're given the function $2x$, you can ask the question what is the function such that when you differentiate it you get $2x$? To answer that question, you would use integration—the inverse process of differentiation: $\int 2x\,dx=x^2+C$ because $(x^2+C)'=2x$.
