Is Integrate antidifferentiate?

Question

I'm still introducing myself in derivation, antiderivation and integrals, because I haven't studied it in class yet (I will study it next year and derivation in some months), but well, about my question I have seen that integrals and antiderivatives, at least, looks like the same, so, is that true? and also I didn't understand at all why the integration gives you the area, I mean, realize that derivation gives you the value of the slope is easy, but why the antiderivation gives you the area is a bit harder for me to understand, thanks!

Answer 1

There are two types of integrals:

• Definite integrals, which are more-or-less just straight-out defined to be the area under a curve,

and

• Indefinite integrals, which are exactly the same thing as anti-derivatives, but we gave them a second name.

So, why are there two different things with remarkably similar names? Because there's this amazing discovery, called The Fundamental Theorem Of Calculus, that shows that they are damn near the same thing. So what you'll want to do is understand why the FTC is true. But it's brilliant and not at all obvious, so give yourself some time to reach a full understanding of it.

(It's worth noting that all mathematicians have fully absorbed the FTC, and so they often aren't very clear about which type of integral they are taking about at any particular moment, as the FTC lets them switch back and forth. It makes things easier for them, but can make things harder for students who are just learning calc. I apologize in advance for any confusion you personally may experience because of this.)

Answer 2

The following is a rough illustration to demonstrate why the antiderivative of a function gives the area.

Let $f: \mathbb{R} \to \mathbb{R}$ and $F: \mathbb{R} \to \mathbb{R}$ be positive, differentiable functions such that $F(x)$ is the area between the graph of $f$ and the $x$-axis in the interval $(0,x)$.

Consider that $F(x + \Delta) - F(x)$ is the area between the graph of $f$ and the $x$-axis in the interval $(x,x+\Delta)$. Now, for small $\Delta$ we have $F(x + \Delta) - F(x) \approx \Delta f(x)$. Thus,

$$ \frac{1}{\Delta} \cdot \Big[ F(x + \Delta) - F(x) \Big] \approx \Delta f(x) \cdot \frac{1}{\Delta} $$

$$ \frac{F(x + \Delta) - F(x)}{\Delta} \approx f(x) $$

and as $\Delta \to 0$ we have $\frac{F(x + \Delta) - F(x)}{\Delta} \to f(x)$. This latter point is more obvious when you consider $F(x + \Delta) - F(x) \to \Delta f(x)$ as $\Delta \to 0$. So, in the limit we have

$$ \lim_{\Delta \to 0} \frac{F(x + \Delta) - F(x)}{\Delta} = \lim_{\Delta \to 0} f(x) = f(x) $$

Notice the far left hand side of the equation satisfies the definition of a derivative. Thus,

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

Hence, by definition of derivative, $f$ is the derivative of the function $F$ that conveys the area between $f$ and the $x$-axis. Or in other words, the antiderivative of $f$ is a function $F$ that conveys the area between $f$ and the $x$-axis.