3
$\begingroup$

I've only recently learned calculus and I am struggling to understand why integrals give the result that they do. I have watched 3blue1brown's videos which I have found extremely helpful in understanding as much as I do.

I understand that (indefinite) integrals are the antiderivative, however, there should be some way to calculate them while thinking of them as the area under the curve without just thinking of the inverse derivative, right?

My intuition to this point leads me to believe that the following is on the path to this way of thinking:

  1. Slice up the area under the curve into many rectangles with widths of $dx$
  2. The height of each rectangle is $f(x)$ at that point
  3. The total amount of rectangles from $0$ to some point $x$ is $\frac{x}{dx}$
  4. Therefore, as a function of $x$ where $F(x)$ = the sum of the area under the curve from $0$ to the point $(x, f(x))$, $F(x)$ = [area of rectangles]$\cdot$[amount of rectangles] $= dx\cdot$[average height]$\cdot\frac{x}{dx}$
  5. This can be simplified to $F(x) = x\cdot$[average value of $f(x)$ from $0$ to $x$]

My problem is that I have no idea how to find the average value of $f(x)$ for any general function, or for any general polynomial. And I can't think of another way to imagine the integral. By intuition, I would say that the average value of $f(x)$ for $f(x) = x$ is $\frac x2$, and it makes sense for the integral of $x$ to be $\frac{x^2}{2}$ because the derivative of $\frac{x^2}{2}$ is $x$. But I don't know why, through intuition, the average value of $f(x)$ for $f(x) = x^2$ should be $\frac{x^2}{3}$ or etc.

$\endgroup$
4
  • $\begingroup$ Taking $t=ux$, $\int_0^xt^{n-1}dt=x^n\int_0^1u^{n-1}du$. If Alice is one of $n$ players in a "whoever has the biggest random number from $0$ to $1$ wins" game, her chance is $1/n$; this must be the average of $u^{n-1}$, her chance if her number is $u$. $\endgroup$
    – J.G.
    Commented Aug 19, 2023 at 16:58
  • $\begingroup$ You said you watched 3blue1brown's videos, but also that you didn't have intuition about why the integral of $x^2$ is $x^3/3$. Is there something about his "volume of a pyramid" explanation of that fact you wanted clarification on? $\endgroup$
    – Mark S.
    Commented Aug 19, 2023 at 17:04
  • $\begingroup$ @MarkS. I watched his playlist on the essence of calculus, I didn't see a volume of a pyramid explanation. Can you direct me to it? $\endgroup$
    – LnlyCloud
    Commented Aug 20, 2023 at 0:21
  • $\begingroup$ I was sure it was him, but I was mistaken. I couldn't find the original video I had in mind, but here is a new video showing the idea I was alluding to: youtu.be/DRzmT-E0ncQ $\endgroup$
    – Mark S.
    Commented Sep 11, 2023 at 23:20

2 Answers 2

1
$\begingroup$

When you average consecutive squares, e.g., $1$, $4$, $9$, $16$, $25$, notice that the squares are not evenly spaced, i.e., $9$ is closer to $1$ than to $25$. So that should give you some initial intuition as to why the average of $x^2$ over is $[0,x_0]$ is not merely $x_0^2/2$, and should actually be a bit lower, since the values of the squares are skewed towards the bottom in some sense.

As a general matter, rather than using averages to compute integrals, we often work in the reverse - using the integral to determine the average through the formula that the average of $f$ over an interval $[a,b]$ is $$\frac{\int_a^bf(x)dx}{b-a}\text{.}$$

Generally speaking, if you want intuition for calculating integrals that relates to areas, consider the following (which will also explain the connection with derivatives):

Instead of trying to multiply the average area of a rectangle times the amount of rectangles, imagine you are simply summing the area of rectangles $f(x)\times dx$. Then ask yourself, what is average rate of change of this sum, as I add one more rectangle? The change in the output sum is the area of a new rectangle $f(x)\times dx$, while the change in the input is merely $dx$. Thus the rate of change, or change in output over change in input, is $\frac{f(x)dx}{dx}=f(x)$.

This is the intuition for the Fundamental Theorem of Calculus and the relation between the derivative (i.e., rate of change) and integral: the rate of change of the integral should be the function itself. We can formalize the Fundamental Theorem (what I call part II but others call part I), by saying that $$\frac{d}{dx}\int_a^xf(t)dt = f(x)\text{.}$$

(You can make all of this discussion much more precise via Riemann sums, and many many texts and online resources will do this, but sometimes the notation to make things precise can be overwhelming and obscure the main idea.)

$\endgroup$
2
  • $\begingroup$ Thank you! That actually did help me understand a bit better why the derivative of the integral is the original function. But I still think there should be some way to visualise it better. Experimentally I have tested the sums of $x^2, x^3$ divided by number of terms and found that they are $\frac {x^2}{3}$ and $\frac{x^3}{4}$. Do you know of a reason why it's specifically those numbers, without using integrals to find them? $\endgroup$
    – LnlyCloud
    Commented Aug 20, 2023 at 0:56
  • $\begingroup$ Every way that I know for deriving these integrals involves either using the derivative power rule along with the fundamental theorem of calculus, or more directly using summation formulas for summing consecutive $n$th powers. I.E., there are formulas that tell you, for example, that $1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$, which can let you directly compute an integral without an anti-derivative. But those are, needless to say, a little more advanced and not necessarily so great for intuition either. $\endgroup$
    – M W
    Commented Aug 20, 2023 at 2:20
0
$\begingroup$

There is no good "average" value. You have either to do the summation for n rectangles and then consider the result for n to infinity, but then you have to know formulas for sums. Or you look at the proof, that the derivative of the "area" function is the function you want to integrate. So you Kow its the antiderivative,

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .