Math: Calculus Video #9

3Blue1Brown Intro To Calculus Series Summary

Carlos Manuel Jarquín Sánchez

9 min readSep 2, 2023

Preface ✨

Hello everyone! My intentions for writing these articles are:

Explain technical knowledge about geology in simple terms (to the public)
Document my journey to getting an AD-T Geology degree in one year
Store information and habits for my future self and others (in <7 minutes)

Coolio? Sweet. Enjoy the series :-)

notes as i watched the video 📹

p.s. — the video will be linked at the bottom of the article. cheers.

from the previous video, we began proving what calculus was built off of:

finding values and significance on a specific point in space & time (on a curve)
integrating points on specific endpoints to find an area under a curve.

but if u ever wondered this:

what problems would we have to use integrals in real life?

well, one of them include finding the average value of a function (given two specific endpoints)?

finding the average of a continous variable (like f(x)) allows us humans to explain why derivatives & integrals are the inverse of each other.

here’s a more practical example using wave functions, (like the ones ur phone uses to communicate with ur friends/fam):

what is the average value of sin(x) between 0 & π?

we’re looking at where the line is yellow, and the area is between the x-axis and the border of the curve.

in simpler terms:

what is the average value of the shaded yellow line?

…but how do we find the average of a continous function (aka never stops)?

with average values, we can do this:

step 1: calculate up a finite amount of values with a specific number value (i.e. 100 points, and one of those points had a value of 27)
step 2: add these finite values together
step 3: divide it by the finite number of points used to make the sum (i.e. in this example, 100)

we picked 4 points, added the values up, and got an asnwer.
we used 4 numbers (points) to get 12.
so divide the answer by the number of points used to get 12 (in this case, it’s four)
simplified, the average of these values between one and five (1,2,4,5) is three.

but that works when the values stop somewhere.

but the graph of sin(x) has no limit to it.

it can go from negative infinity to infinity forever!!

so that little area between 0 & π has an infinite number of values to add up!

but if there’s an infinite values, that means we would need to divide by infinite values… which cannot exist.

**each yellow line is one point (number value). but now, we have infinite points, especially if the distance between each yellow line is almost zero.**

and the infinite points cannot be divided by infinity. that number is not quantifiable. (i.e. u cannot hold infinity number of marbles in your hand…’cause what IS infinity marbles in ur hand, rly?)

and tbh, this occurs a lot in math…

especially for finding the average of infinite points…

you want to add them up, divide, and find the average number of fixed endpoints.. but for infinite points, it don’t make sense.

and this is when…

sure… but how do we fix this mess?

step 1:

approximate the sin(x) from 0 to π with a finite sum… kinda.

here, lemme show ya.

imagine sampling a finite number of points between sin(x) from 0 to π.

(and yes, these finite points have values, like sin(π/2) or sin(π/4).

step 2:

add up all these finite points from sin(x) from 0 to π.

step 3:

divide these sample points by the same number of samples

**(i.e. if u had 50 samples points, u divide by 50)**

HOWEVER!

the more (hypothetical) finite points we add to the numerator…

the closer the average of that sample should be to the TRUE average of the continuous variable…

and see that?

this looks like a job for the integral.

if we’re going to add up an infinite number of points,

we must add all of these points up with the integral symbol “∫”…

ESPECIALLY when a “dx” is present…

because the “dx” (aka distance from original point to changed point) affects the width between each number we measure.

also, “dx” is the integral used to calculate the average of all the numbers/average points.

and a continous variable (for a continous function) will look like this for sin(x) between 0 to π:

we measure the area of the yellow, shaded curve by sin(x) * dx

and yes, the integral is used to represent as “dx” approaches zero.

from going from a fraction to an integral… this is what it means in english:

fractions: sum of heights (with a number/variable) divided by the number of sampled points…

TO…

integral: the spacing between each sampled point (aka, the “dx”)

example time.

if the spacing between each sampled point was “0.1” from 0 to π…

how many points can exist from 0 to π on the sin(x) graph IF the spacing between each point is 0.1?

step 1:

take the length of the interval.

in this case, it’s π, or 3.14…

how did i find this?
the starting point was 0, and the end point was π from the shaded region on the sin(x) graph.

so we subtract the endpoint from the starting point (bigger number goes first).
so, π — 0 = π units in length.

step 2:

divide the interval length by the point spacing. there, we’ll get how many points can exist…

the interval length is π.

the spacing was 0.1

divide: ||| π / 0.1 = 31.4159265359

simplify it, and it’s 31.

the “31” means that 31 points can exist between sin(x) from 0 to π IF the point spacing between each line was 0.1

BUT!

here is the catch between all of this.

the “point spacing” is technically “dx”.

because “dx” IS the spacing between each sampled point.

so we can replace with “0.1” with “dx”.

before: π / 0.1 ||| after: π / dx

so all those samples, divided by (π/dx), becomes…

it’s all the points, times the spacing between them, all divided by π.

but if we’re being honest, the numerator is just an integral,

and the integral looks like this:

the integral starts at the value where x = 0, and ends at the value where x = π.

we’re adding all the points between 0 to π on a sine wave,

and multiplying those points by the point spacings between each value (dx).

P.S. -) the more samples we have for larger & larger samples of points…

the average approaches the actual integral below:

The average height of the sine wave function between 0 & π…

IS the area divided by its width.

the shaded part (colored in green-ish) is the area.

the width is the distance 0 to π on the number line.

ok, now to solve this gunk.

solving ∫ [(π)(0) sin(x) dx] / π → for average height

first, what is the anti-derivative of sin(x)?

well, if u know, the derivative of sin(x) is cos(x).

the anti-derivative of cos(x) is sin(x).

so the anti-derivative of sin(x) is -cos(x).

next, we have our anti-derivative.

so we now must plug in our starting points & endpoints into the anti-derivative to find the average height of the graph.

and that’s in the equation above.

[cos(π)] is -[-1], or just 1.
cos(0) is +1.
so 1+1 = 2.

now, we must divide 2 by π.

and our answer to the average height of a sine wave between 0 & π is…

2/π

but hold up.

the height of our anti-derivative is 2, (’cause we add the total distance from -1 to +1, which is 2 units in distance.)

and the distance from the bottom of the -cos(x) wave to the top takes π units to get there, that’s the width of the wave.

that’s why it’s 2/π.

btw, 2/π is about 0.64 units. (2/3.14 ≈ 0.64)

IT ALL COMES BACK DOWN TO THE RISE OVER RUN OF A GRAPH LMAO.

HMMM!

so the average slope of a graph (over all its point in a certain range) must equal the total slope between the start & the point?*

* translated from english to math, i mean this:

(Y2 — Y1) / (X2 — X1)

and visually, it would look something like this:

and due to the endpoints, our complex problem of finding the average height/distance has been distilled into two simple steps:

for any function “f(x)” between intervals (a,b), do this…

step 1:

take the integral of F(x) on the (a,b) endpoints ||| aka any value inside of the (a,b) integral
divide the integral over the width (aka distance) of that interval (a,b)