First year college CALCULUS and PHYSICS, I can teach. Send PM

[Nominal Animal]

Nominal Animal:
   ? Would you be able to to take a couple seconds, to explain something you outlined ?

   It's;       (X1 - X min.) and that was multiplied by two;   W1 = 2 (X1 - X min.)

   I didn't follow that, as to where the '2' came from, in the Rieman method.  Plus, actually, not making progress putting that (above formula) in a plain English and graphical context.
   Help appreciated !

Rick

Good catch!  It should be *divided* by two, not multiplied by two.

Let's consider the case where we have five \$x\$ coordinates, \$x_\min\$, \$x_1 = -1\$, \$x_2 = 0\$,  \$x_3 = 1\$, and \$x_\max = 2\$,
in the middle of each interval, each rectangle –– except for the first and last one –– so we have five intervals in total.

The first interval is from \$x_\min\$ to midway between \$x_\min\$ and \$x_1\$: \$x_\min\$ to \$(x_1 + x_\min) / 2\$.
The second one is from midway between \$x_\min\$ and \$x_1\$ to midway between \$x_1\$ and \$x_2\$: \$(x_1 + x_\min)/2\$ to \$(x_2 + x_1)/2\$.
The third one is from midway between \$x_1\$ and \$x_2\$ to midway between \$x_2\$ and \$x_3\$: \$(x_2 + x_1)/2\$ to \$(x_3 + x_2)/2\$.
The fourth one is from midway between \$x_2\$ and \$x_3\$ to midway between \$x_3\$ and \$x_\max\$: \$(x_3 + x_2)/2\$ to \$(x_\max + x_3)/2\$.
The last one is from midway between \$x_3\$ and \$x_\max\$ to $x_\max\$: $(x_3 + x_\max)/2\$ to \$x_\max\$.

Their widths are thus
$$\begin{aligned}
w_1 &= \frac{x_1 + x_\min}{2} - x_\min &= \frac{x_1 - x_\min}{2} \\
w_2 &= \frac{x_2 + x_1}{2} - \frac{x_1 + x_\min}{2} &= \frac{x_2 - x_\min}{2} \\
w_3 &= \frac{x_3 + x_2}{2} - \frac{x_2 + x_1}{2} &= \frac{x_3 - x_1}{2} \\
w_4 &= \frac{x_\max + x_3}{2} - \frac{x_3 + x_2}{2} &= \frac{x_\max - x_2}{2} \\
w_5 &= x_\max - \frac{x_\max - x_3}{2} &= \frac{x_\max - x_3}{2} \\
\end{aligned}$$
The height of the first rectangle is \$f(x_\min)\$, the second rectangle \$f(x_1)\$, the third rectangle \$f(x_2)\$, the fourth rectangle \$f(x_3)\$, and the fifth rectangle \$f(x_\max)\$.
Thus, the areas of each rectangle, i.e. the approximate integral over each interval, are
$$\begin{aligned}
A_1 &= f(x_\min) \frac{x_1 - x_\min}{2} \\
A_2 &= f(x_1) \frac{x_2 - x_\min}{2} \\
A_3 &= f(x_2) \frac{x_3 - x_1}{2} \\
A_4 &= f(x_3) \frac{x_\max - x_2}{2} \\
A_5 &= f(x_\max) \frac{x_\max - x_3}{2} \\
\end{aligned}$$
and the approximate integral is their sum, \$A_1 + A_2 + A_3 + A_4 + A_5\$.

Yes, the first and last rectangles are narrower than the others.  (I also should have shown how \$w_2\$ and \$w_{N-1}\$ are calculated, but forgot, because I usually use \$x_0 = x_\min\$ and \$x_{N-1} = x_\max\$.)

[RJSV]

   Actually, maybe I 'caught' one small glitch...

   However, feels like the fictitious Rip Van Winkle story, where, this time,  I've woken up and discovered a whole new world (mathematics):   it's maybe like a fabric with many new moth holes..., or actually mostly holes with nothing but air, plus a few strands of intact fabric.
   Just trying to get a more literal grasp of the nomenclature, for labels on repeated divisions (using 'i').  How to work in the time variable 't' in those, as it's differential with respect to 't', and integrals w. respect to 't'...Getting all that coordinated, before even dealing with solving or modifying (the equations).
   I don't recall an explicit use of 'anti-derivative' anywhere, although certainly it was discussed in classroom...(that was what? 55 years ago!)

Anybody see, where I've set my coffee cup ?
- - -Rick

[IanB]

I don't recall an explicit use of 'anti-derivative' anywhere, although certainly it was discussed in classroom...(that was what? 55 years ago!)

You would use the anti-derivative when doing symbolic integration. For instance, if you want to integrate \$x^2\$, you will need that the anti-derivative of \$x^2\$ is \$\frac13 x^3\$. That will give you the indefinite integral as:
$$\int x^2 \,\textrm{d}x=\textstyle\frac13 x^3 + C$$
Or the definite integral as:
$$\int_a^b x^2 \, \textrm{d}x=\left[ \textstyle\frac13 x^3 \right]_a^b$$

[RJSV]

(December 5th post has my diagram, of the three hierarchial related formulas, increased 'order' equation has X squared, mid equation at complexity of 'X', and lastly the third, least complex equation is just a constant.
   But, again, don't recall anything said about 'anti-derivatives', although that dynamic (of reversal) was certainly a major part of explanation of the history of developing an ever expanding table of integrals (solutions).

   Also, similar comment for the phrases 'indefinite' and 'definite' that are associated with anti-integration.  Instructors maybe used different phrases, in their classroom.

[IanB]

You have a mistake in your formulas from that December 5th diagram that would lose you marks as a student.

For example, if the (constant) acceleration is \$a\$, then the velocity is given by
$$v=at+C$$
where \$C\$ is a constant of integration. \$C\$ can take any value, but is commonly chosen to be \$v_0\$, the initial velocity at time zero. Thus we would have:
$$v=v_0+at$$
Similar considerations apply when integrating velocity to get position.

The terms "definite integral" and "indefinite integral" are, I think, universal in time and place. Perhaps you just don't remember them?

The idea of a definite integral would apply to the change in velocity when integrating the acceleration. Thus, we could say:
$$\Delta v=at$$
In this case the change in velocity is definitely known, and therefore there is no constant of integration. An indefinite actual velocity became a definite change in velocity.

[RJSV]

Yeah, NO, I'm pretty sure I'd remember that, but remember, classroom tests etc were not well focussed all the way to a numerical answer, such as distance traveled, water amounts leaked, etc.  More like proofs and theorms.

   Quiz questions were more like area under such and such curve, for some of that, but more like show what is needed to calculate that.

  (Maybe that reflects the traditional 'white collar' engineering in office setting, meanwhile the technicians ran around obtaining the numerical results...in the lab or on factory floor)

[IanB]

Well the area under a curve is an example of a definite integral. For example, if we say that the area under the curve of y = x² between x = 3 and x = 6 is 63, then that is a definite integral. Any time the answer can be stated as an actual number, that is a definite integral. If the answer contains a constant of integration, and we don't know the value of the constant, then that is an indefinite integral.

[RJSV]

   Beginners reading, here's an outline, of what's happened, (in some simple language):
   It was pointed out that the expression for the speed of an object is:
     'a X t'  plus  a constant, and that constant would be your speed at an earlier time, when you started.  For example, moving at 4 meters per second, at start of some time interval.  That's the constant. 
   Then, the 'a X t' term gives the accelerated motion or what's called free-fall, from the start of the time interval, on to an end (of measured time period).  Let's say the object picked up 10 meters per second...that means you end up at speed of '4 + 10' or 14 meters per second.

   I believe you have similar situation for calculations of distance.  That is, integrating the speed term, above, which was 'at + c' would give the equation '1/2 X a t squared' plus 'ct', and then plus another constant, (let's call it c2).

   That 'c2' comes from integrating as explained by Ian, and that 'ct' portion of your position gives gives a way to account for whatever position is when an interval is started (its speed x time)= position.)

   Lol, this is making me squint really hard, but simple explanation is that you are accounting for the initial position, and initial velocity, in your measure interval, and then calculating the ending position.
An initial position and initial velocity, if there, and plus the changes during the (time) interval gives you the real position, at end.

....whew, complicated to explain, in proper frame.!

[RJSV]

Yes, thanks Ian, for the constructive criticism (reply #79 from Nov 28th).  Pushing on those comments took a bit of time.
   But what I get, when attempting to do the integration, for POSITION, is three terms.  That was a messed up area, as there were 2 constants, or so it seemed, after integrating from the velocity equation:
   One of those 'constants' gets re-labled as V0, the initial velocity, for the current measurement interval, while the other constant is simply the initial position.  That's nice, as I didn't like the look, with having TWO constants (after integrating the Velocity line equation).

 I got:   (1/2 a X t squared) + (V0 X t) + C
...where the 'C' is your initial position, or y0, (assuming its motion along y axis).

   This should make for a nicer stack of the three equations (Position / Speed / Acceleration) in the derivative / integration heirarchy diagram.
Thanks a lot, Ian!

[RJSV]

   So in getting down to business, with a new year and all (the holiday parties done), I've come up with a count of 16 places that you folks have responded in detail...usually with extensive formulas illustrated!

   That's kind of high volume of work, to wade through and figure through (most of them anyway).  If I can average approx. one every two days...brings me out, into the light around February ...plus a week or two.
   That's some (moderately) dense shit, (and intimidating)!

   But of course I could unpack a good deal of it by way of asking questions, of the handful of maybe four members who posted, over November and December.
   Ambitious, but if done transparently I might actually be able to learn a thing or two, while other members learn the stuff.

[RJSV]

   I was saying that there are 16 value-packed responses, and since I can't easily generate a decently formatted TABLE (lame cellphone input), I'll just list 4 entries, for now:

   (helpful) replies #29, #46, and #47 (by IanB),
   and reply #50; (Mr.AI).
   Plus you can see the 'hierarchy' diagram,
reply #27. Up

   Some of difficulty I'm having is simple nomenclature stuff, not too hard, but initially confusing.  I was looking for a 'slope' in graphical sense, to be 'y/x' or 'dy/dx' in the geometric sense, but if you use the cannonball parabolic path, (reply #49), can start correcting the nomenclature.

   The 'Independent variable' is horizontal, that would be the elapsed time (not 'X axis that I usually would consider).  We have to ignore air friction, so that horizontal time axis is evenly divided.

   Next, the ACTUAL use of 'X', is as the dependent variable, for the cannon shot problem, confusing, simply because that 'X' variable is VERTICAL, on the graph, and, is backwards in some gravity problems.
(I'm leaving that go for a second, as a cannon ball parabola has both up and down travel acceleration and deceleration).

   O.k. so, the main differential will actually be
   dx/dt, as the projectile moves up, and then down.  Contributer IanB has this path and distance problem with shorter path on the right half.
   That gets the first stuff into more coherent status.  Now to the integration hierarchy,
(reply #24).  While I could understand the first integral, taking the acceleration constant and creating the two items of speed, doing that once more seems to then end up with TWO constants;  That's been explained now, as you simply need to re-lable your velocity starting point (constant), to 'u'

   That way, when doing the second integration, (reply #24), you get, in general, THREE terms, which I like to classify as:
  (Top) (1.) Acceleration Constant.
  (Middle) (2.). Velocity linear+constant.
and finally, (bottom formula) (#3.) The full Squared term, linear term and constant, defining you position.

   It makes good sense, and a 'neat' form that has the more 'generic' quality in the relations, of the three equations for a 'free fall' type of (cannonball) motion.

   Nice so far, but needs some minor considerations for the SIGN, positive or negative, of the gravity caused motion.  I guess it's really too much to ask, to do these aspects without a quite literal feeling for the actual case or problem posed.

Thanks

[RJSV]

   One roommate here wants me to 'derive' the equation for the area of a circle, 'π r exp 2', (the classic Pi R squared).
   Basic formula is
 r squared = X squared + Y squared,
in various forms, but I think I'd start from a function standpoint:  y = f(x).

   Looking at my included diagram, I've come up with an interesting take on the various option routes for solving the area, that I wish to share:

   Viewing the included diagram, I considered obtaining an integration solution, for area under a subsection of a full disk, or even by way of Angular function, using infinitesimal angle segments.
   Thinking, If you do the integration within 1/8th of the disk, that would lead to the complete result, by way of multiplying by 8.
   Now, when using y= f(X) you could always perform the area calculation arriving at a four sided shape that has a curved top, using X value that ends aligned with 45 degree point...that would be 0.707 X the full x at radius size...0.707 X radius.
'Cute' thing is that it's then easy to turn the equation into the area of 1 'pie slice' shape, by simply subtracting the triangle area left, by the integration, that is the trianglular area underneath the pie-shaped area.  Nice.

   However, it's likely easier to do integration over the entire quarter circle, as appears in the first quadrant of disk.

[IanB]

It's easier to consider thin "shells" around the circumference of the circle and add them all up.

For instance, the circumference of a circle is 2πr. Then the area of a small "shell" of thickness dr around the circumference is 2πr dr

Therefore the area of a circle of radius R, say, can be found by adding up (integrating) over all the little shells:

A = ∫ 2πr dr between 0 and R ( = πR² )

[IanB]

The area can equivalently be found by integrating in \$xy\$ coordinates, but it's much more complicated, as you can see below.

We have for the equation of a circle:
$$x^2+y^2=r^2$$
We can find the area of the quarter circle in the upper right quadrant with this integral:
$$A=\int_0^r y\,\mathrm{d}x$$
Substituting for \$y\$:
$$A=\int_0^r \sqrt{r^2-x^2}\,\mathrm{d}x$$
To integrate this, we can make the substitution \$x=r\sin{t}\$, whereupon \$\mathrm{d}x=r\cos{t}\,\mathrm{d}t\$
With this substitution:
$$A=\int_0^\frac{\pi}2 \left(\sqrt{r^2-r^2 \sin^2{t}}\right)\left(r\cos{t}\right)\,\mathrm{d}t$$
$$A=\int_0^\frac{\pi}2 \left(\sqrt{r^2(1- \sin^2{t})}\right)\left(r\cos{t}\right)\,\mathrm{d}t$$
$$A=\int_0^\frac{\pi}2 \left(\sqrt{r^2\cos^2{t}}\right)\left(r\cos{t}\right)\,\mathrm{d}t$$
$$A=\int_0^\frac{\pi}2 \left(r\cos{t}\right)\left(r\cos{t}\right)\,\mathrm{d}t$$
$$A=r^2 \int_0^\frac{\pi}2 \cos^2{t}\,\mathrm{d}t$$
To make the next step we can use the identity \$\cos{2t}=2\cos^2{t}-1\$. Thus:
$$A=\frac{r^2}2 \int_0^\frac{\pi}2 \cos{2t}+1\,\mathrm{d}t$$
Solving the integral gives:
$$A=\frac{r^2}2 \left[\frac12 \sin{2t} + t\right]_0^\frac{\pi}2$$
$$A=\frac{r^2}2\left(\frac{\pi}2 \right)=\frac{\pi r^2}4$$
Since this is the area of a quarter circle, the area of a whole circle is four times this, or:
$$A=\pi r^2$$

[RJSV]

   Thank you, this is very much what I was thinking, (except I'm more interested in getting a 1/8th portion, as illustrated).
   Doing the 1/8 'pie slice' intersecting the perimeter at 45° (degrees), the triangular area to subtract becomes 0.707 X 0.707, which in perfect numbers will give exactly 1/2 square inches or square centimeters, depending on using a 'unit circle' of radius = 1.
The reason I like to play with this alternative is, mainly that I haven't seen that before...just exploring...(out of love, for pure mathematics)!

   Of course it's much more difficult, but I think using that 'y = f(X)' form is more formal and general.  But now, the sort of reverse proof is to differentiate the integral...that is using the 'rote' or memorized way;
   Taking the exponent (2) down into the multiplying terms, getting '2π' then, and decreasing the actual exponent, to R to the first power, gets your anti-integral, '2πR'.
That goes back and forth, if you want, between Integral and accompanying derivative, as I've mentioned being partially using 'rote' (memorized) and the reversal...if there is one, in your bag of tricks (table of solutions).
   But then, of course, question comes to mind, is there an actual literal method, other than;
   "I've seen that one, where a constant times a variable ends up as constant times the variable squared".
In other words, the 'rote' method, (and it's anti-calculation) seems a little like cheating, with some luck....Is there a more defined way, of building up an Integral like this, rather than the method of: "Saw that solution before" ?
   Sorry if my English structure is a bit circular (no pun).
   At any rate, after obtaining the full area, under the curve, from 0 to 45° that triangular region comes out to a half square, of 1 unit by 1 unit (1/2 of that).
Thanks

[IanB]

   Thank you, this is very much what I was thinking, (except I'm more interested in getting a 1/8th portion, as illustrated).
   Doing the 1/8 'pie slice' intersecting the perimeter at 45° (degrees), the triangular area to subtract becomes 0.707 X 0.707, which in perfect numbers will give exactly 1/2 square inches or square centimeters, depending on using a 'unit circle' of radius = 1.

Surely it's 0.5 x 0.707 x 0.707 = 0.25? (Area of a triangle is half base times height.)

The reason I like to play with this alternative is, mainly that I haven't seen that before...just exploring...(out of love, for pure mathematics)!

Although the triangular bit may be "easy", you still have the circular bit to deal with afterwards, which will use exactly the same method I outlined. So it's really no easier at all.

Taking the exponent (2) down into the multiplying terms, getting '2π' then, and decreasing the actual exponent, to R to the first power, gets your anti-integral, '2πR'.
That goes back and forth, if you want, between Integral and accompanying derivative, as I've mentioned being partially using 'rote' (memorized) and the reversal...if there is one, in your bag of tricks (table of solutions).
   But then, of course, question comes to mind, is there an actual literal method, other than;
   "I've seen that one, where a constant times a variable ends up as constant times the variable squared".
In other words, the 'rote' method, (and it's anti-calculation) seems a little like cheating, with some luck....Is there a more defined way, of building up an Integral like this, rather than the method of: "Saw that solution before" ?

Of course, there is naturally a literal method. Nothing in mathematics is "rote". It all can (and must) be derived and proved. Any reasonable calculus course will include these derivations and proofs. I learned them when I was 16 and can still broadly remember them (or can recreate them). What I did above was pulling back memories from high school mathematics studies about 45 years ago.

You will see that I didn't use a "table of solutions" to do the integration in my post above (though I could have). Rather, I showed the actual details. It was easier to work it out from scratch than to spend time looking for a table of integrals.

[RJSV]

   Oops, yeah, I goofed that one up, estimating the area, but off by factor of two!  Sorry.
I think I got distracted, into thinking about a shortcut to a smaller portion of the basic multiplication tables.

   One shortcut involves the multiplication of 1 digit, below 10, by another digit above 10.
Doing this (see diagram), gives a kind of 'reflection' with the '10' in the middle, of a number line.

[RJSV]

   So, it's a simple math related trick, and uses a Quality Factor, usually up as close to 100 'perfect'.
   Taking 9, 8, 7, and 6, while the lower multiplies are too trivial to deal with),
You will have; 9 matched or paired with 11.
That, multiplies as '9 X 11' or 99, thus having a Quality Factor of '1', or minus one, technically.  But it's the distance from '100'.

   Doing the '9 X 11' quick multiply, the more accurate version will be, (interestingly);
   '9.090 909 09...etc.' X 11, to get perfect 100 as result.   OR, you could do it as;
   '9 X 11.111 11 repeating' as giving a perfect result.

   Doing this, as pairing '8' with a '12' gives slightly off-results, of 96 instead of 100.

[RJSV]

   Now you can do this little memory speedup for just a few;  '9, 8, 7, and 6, practically.  The case for 7 is also interesting;   When doing
'7 X 14' the Quality Factor is good, at 99, and there is (shown in previous diagram) the wonderous result of '0 .142857142...' as mentioned elsewhere, as a longer pattern of 6 digits that repeats, in the fraction.
But also, there is the '7' doubling itself, at least briefly, doing '7, 14, 28, and 57,... ALMOST continuing the simple digit doubling.
The pattern occurs, similarly when doing division, say of 1/7...   Some of that seems to veer (dangerously ?) close to the simple √2.

[IanB]

It's a common mental multiplication trick.

In general, you have:
$$(x-a)(x+b)=x^2-ax+bx-ab$$
So, if you wanted to multiply 99 by 102, you would have:
$$(99)(102)=100^2-100+200-2=10,098$$
Taking your example of 8 times 12, we have:
$$( 8 )(12)=10^2-20+20-4=96$$
More interestingly, 98 times 102 gives:
$$(98)(102)=100^2-200+200-4=9,996$$

Also, 1/7 being 0.14285714... is not so much close to √2 as to π. Hence we have:
$$\pi\approx\frac{22}{7}$$

[RJSV]

Hey IanB ! Thanks for reading!

   You get a more interesting figure, when dividing by 14, instead of 7:
   1/14 = 0 .071428571...

Note that it has the same structure, where the pattern should be doing '56' as the next doubling value,...but, alas, it diverges from that pattern (of new lower value digits in the fractional number).
 
   It gets interesting, when trying to compare with 'root two' related values.  That would include the various such as '1/√2'; at '.707106...'

Maybe this has simpler explanation, but also relates to the use of '22/7' in some calculation shortcuts.

   Now looking at the extreme ends, you would 'pair' the numbers 2 and 50;  (for a perfect quality factor).  The next 'pairing', you may notice, would be 3 with 33.333... which is (approx) 17 counting units apart, and each subsequent pair gets a smaller and smaller separation.  If you had started on the other end, it would appear that the distance stepped is exactly '1.0' but that quickly shows as false assumption, (witness the '13' got skipped, in the next pair, '7 with 14'.)
At any rate, by the time you've tracked each pair, down to (4 with 20), it becomes really too trivial to include in some simple gimmick...it's just apparent on inspection, with '4, 3, and 2' doing the in-head figuring.

   Playing with the various combos, for illustration with an example (estimate):  Suppose someone says that business improved by about 1/9th last year, you immediately can be thinking:  'Uhh, that'd be 11 percent gain,...(or multiply by 1.11 for the gained new figure.)

   Another quick example, someone saying 'down by 1/16th,...you know it's (about) 6 percent diminished...and about how accurate your rule of thumb is...
That is, '6 X 16' being a bit 'low' in this context.

[Scott_Z]

Right now there's a 'toss-up' as to which is relevant, (historically in the discovery process), as to whether you show a complete and separate derivation, of each of the derivative, and of the integral.  That is, whether the reversability of the two calculus functions can be enough to get your INTEGRAL, after deriving the derivative.   That would involve building up a set or table of known integrals...when each derivative is determined.
   I find that the 'Summary in plain English' I can do very well (eventually, lol), whereas the exact formula manipulations are a (bulky) challenge.

   Many, many reaaaallly smart doctors and some teachers live in this so-called common sense 'gap' or bubble, often lacking common sense oversight, but meanwhile doing very well in their descriptive math.  It takes just a sheer volume of study time! By the way speaking about colleges and education.  Platforms like Edubirdie's https://edubirdie.com/mba-essay-writing-service MBA Essay Writing Service could complement your academic journey by offering additional support and resources. Keep up the dedication to learning and teaching—it's through persistence and study that mastery is achieved!

That's a good point! When it comes to calculating derivatives and integrals, there are indeed different approaches, and the choice often depends on the context and the specific task at hand. Building a set or table of known integrals can be a useful tool, especially when working with standard functions.