Area under y=x^3 from 0 to 1, Riemann sum vs. Integral Power Rule

[@ln7247]

Too late I already took calc and did horrible.

[@alejandroduque772]

Hi bprp, could you make some videos about multivariable calculus, maybe some limits and tips. Love your vids:)

[@jacoboribilik3253]

Riemann sum is the perfect example of an incredible theoretical accomplishment yet a poor tool for calculating what it was meant to explain.

[@karstenmeinders4844]

Well, the rectangle sum behaves to the integral like a rasp to a CNC machine! A very good example to demonstrate how powerful the integral is even when f(x) is such a simple function.

[@stephenbeck7222]

Have you ever done any problems going from Reimann sums in sigma notation to definite integral notation? I think that’s a good 1st year or AP topic.

[@helloitsme7553]

What you mean is : infinite sum VS antiderivative, since this is integrating as well

[@angelmendez-rivera351]

HelloItsMe Not necessarily. Integration is defined as an inverse operation of differentiation. Summation is merely a method to obtain integrals which works SOMETIMES. Lebesgue integration, for instance, has nothing to do with infinite sums and is fundamentally different, and it typically uses pathological or ill-defined antiderivatives, so a distinction between infinite limit Riemann summation and integration must be kept.

[@helloitsme7553]

@@angelmendez-rivera351 no. Integration literally means calculating areas of graphs. It doesn't necessarily have to do with derivatives at all

[@BigDBrian]

technically in the riemann sum part, you're not calculating the area of each rectangle but the signed area. But as everything's positive, f(x) is positive, so 1/n * (f(n) is also positive.

[@angelmendez-rivera351]

mrBorkD You contradicted yourself. If everything is positive, then he IS calculating the area. Also, the semantic distinction between signed area and absolute area is merely topological. Strictly speaking, the summation is not topology dependent.

[@BigDBrian]

Okay, strictly speaking there's a contradiction, but only if you ignore what I was actually saying. It's just phrasing

[@JB-ym4up]

Use left corner rectangles in the infinte sum to proove the integral is correct by squeeze therom.

[@nimmira]

Imagine a solution without a box ....... MAMMA MIA!

[@roderickwhitehead]

Sometimes it is good to get back to the basics.

[@ryanrussell6922]

hey can you make a video of some solids of revolution problems?

[@benjaminbrady2385]

I think a great video idea would be a general formula for ln(a + bi). I don't know how you would do it personally but I'm sure you can!

[@angelmendez-rivera351]

Benjamin Brady a + bi = re^it, so ln(a + bi) = ln(re^it) = ln(r) + it = ln(a^2 + b^2)/2 +i[atan2(b,a) + 2πn], n is an integer.

[@benjaminbrady2385]

@@angelmendez-rivera351 Sorry, what does the b,a notation mean inside of the arctan?

[@solinmaaroof]

you are my savior

[@mehrdadmohajer3847]

Thanks. Usually as i see your tries, i come up to find out , if any, other solutions . Here using " e - Funktion " is NOT ONLY obvious, BUT ALSO gives the only accurate value for Area measurement. My calculation shows A= 0.35 which is bigger than A= 1/4 = 0.25 . I´d be happy if you investigate it & show the solution for fans. Thanks.

[@angelmendez-rivera351]

Mehrdad Mohajer There is no such a thing as the "e-funktion" in mathematics, and the calculation A = 0.35 is completely incorrect. A = 0.25 is the correct calculation.

[@mehrdadmohajer3847]

@@angelmendez-rivera351 hi. Thanks for the respond. You see in differential equation dy= f´(x). dx ....the part dx is appproaching 0 ( lim x -----> 0) but not equal to 0. Therefor you got some deviation in calculation, for example Area. With EXP. - Funktion ( e as base ) dx is as Zero as possible ( just one point of tangency ) , therefor no deviation in Area. There is the solution to this problem of base e , y= e^x , which is in my opinion not equal to 1/ 4.

[@Gelo2000origami]

Awesome

[@komminilsen3900]

Can you shoutout pewdiepie

[@NeonArtzMotionDesigns]

What a coincidence, were learning that in calc now

[@michaelblack8111]

Such a cool video!!!

[@abdalrhmanhammad3146]

Can you prove reduction formula for sin x Using complix

[@jackkalver4644]

Is it ever easier to use the definition of a definite integral than to use an indefinite integral?

[@kingbeauregard]

Here's a question: can you have complex limits of integration? As a guess, we'd be getting not the area under a curvy line, but the volume under a curvy plane. (I realize that "curvy line" and "curvy plane" are inherent contradictions, but you know what I mean, I think.) Would love to see you tackle that one of these times.

[@angelmendez-rivera351]

kingbeauregard Not quite. You can have complex limits of integration, but this requires defining a contour and then defining the corresponding contour integral, which is the C analogue of a line integral as defined in R^2.

[@kingbeauregard]

@@angelmendez-rivera351 Thanks!! Would love to see an example. Would totally contribute a new set of pens.

[@GameMaster-pz9pw]

I'm not very advanced in math, so I'm just wondering what the area under y=x^3 is used for.

[@khemirimoez8661]

Absolutely nothing kappa

[@alicwz5515]

Nothing, that's just an introduction to the concept of an integral

[@wenhanzhou5826]

If x^3 is representing a velocity curve e.g your velocity cubes every second. Then you can find the distance you travel after one second by calculating the area beneath.

[@helloitsme7553]

You could have a certain shape that is cut out like a cubic equation and you wanna know it's volume then you have to know its base

[@Casey-Jones]

troll alert

[@przemysawkwiatkowski2674]

Actually both methods are the same... Integral is nothing else but infinite sum. :-)

[@angelmendez-rivera351]

Przemyslaw Kwiatkowski No, that is only true for Riemann integration. Integration is a linear operator which is the inverse of differentation.

[@michabbs]

@@angelmendez-rivera351 Indeed, but... what he did here IS Riemann integration. I was referring to the video. :-)

[@atmonatmon2947]

Shoot a video about what is t : a^b = b^a*t

[@angelmendez-rivera351]

ATMON ATMON What is this supposed to be: a^b = b^(a*t), or a^b = (b^a)*t? Please use unambiguous notation.

[@RahulThakur-em2fh]

Plz solve it.... Integrate lim zero to infinity (x^c/c^x)... Plz sir do it

[@Debg91]

Hi bprp! Can you compute the volume under a 2D graph using Monte Carlo?

[@anggaadandiputra8450]

Chen lu

[@blackpenredpen]

Angga Adandi Putra No Chen Lu here tho.

[@yrcmurthy8323]

He's obsessed with Chen lu

[@noobmaster-dm7tu]

isnt the logic behind the definite integral the Riemann method?

[@MohammedAbdullah-mx1vg]

why not use trapeziums instead of rectangles, surely they will approximate area better?

[@angelmendez-rivera351]

Mohammed Abdullah There is no approximation here. The calculation is exact because the limit was evaluated exactly.

[@vivekchowdhury8879]

Can u please solve the equation Sin2x-√3sinx=2?

[@blackpenredpen]

Vivek Chowdhury Yo, change your profile picture first yea?

[@vivekchowdhury8879]

@@blackpenredpen okay Sir :😊

[@vivekchowdhury8879]

Please don't mind :)

[@angelmendez-rivera351]

Your notation is incomprehensible. Is this equation supposed to read as [sin(x)]^2 - sqrt(3)·sin(x) = 2, or as sin(2x) - sqrt[3·sin(x)] = 2, or some combination of the above?

[@vivekchowdhury8879]

@@angelmendez-rivera351 Sin(2x) -( sqrt(3) × sinx)=2

[@marceloguzman646]

Do you have some basics trigonometry videos?:(

[@yrcmurthy8323]

Yes

[@RitikSingh-wc6ps]

Integrate 1/(1+x^5)

[@angelmendez-rivera351]

Ritik Singh Factor 1 + x^5 into irreducible quadratic monomials, and perform partial fraction decomposition. Use the linearity to have the linear combination of the individual integrals of the reciprocals of the quadratic factors. Then integrate the quadratic factors by completing the square, using substitution, and using the fact (d/dx)arctan(x) = 1/(1 + x^2).

[@mister_allmond]

Wait so am I the only one who has a hard time understanding what this man is saying? Great content nonetheless

[@KnakuanaRka]

I don’t remember hearing anyone else have trouble with his accent, but whatever. If you still love his material, great!

[@yashprajapati8857]

The inverse function is equal to the integral of the function.... Does any function as such exist???

[@jacoboribilik3253]

hmmm it looks like an integral equation.

[@sergey1519]

No.

[@MrJloa]

Hm... The funny thing is that integral is actually a sum of Osmall rectangles :-)

[@btdpro752]

Hi

[@khemirimoez8661]

U play BTD?

[@btdpro752]

@@khemirimoez8661 Sometimes

[@samiunalimsaadofficial]

Hi bprp

[@priyanksisodia5889]

Please, my question is, why limit x tends to 0,log x minus one over x is 1, please explain

[@brainlessbot3699]

The graphs of y=log(1+x) and y=x are very close to each other at x=0.Therefore for small x the quotient log( x+1 )/x is very close to 1 for small x.Actually the lim x-->0 e^x-1/x is 1 and not log(x)-1/x.

[@angelmendez-rivera351]

Mohini hazra There is a better explanation that can use analytically (without requring graphs) and can be used in an exam. Notice that log(1 + x)/x = log[(1 + x)^(1/x)] by the exponentiation property of logarithms. Furthermore, since the logarithm is a continuous function, which can be proven analytically and algebraically using only elementary manipulations, it so happens that (lim x -> 0)(log[(1 + x)^(1/x)]) = log[(lim x -> 0) (1 + x)^(1/x)]. By definition, (lim x -> 0) (1 + x)^(1/x) = e, & log(e) = 1.

[@brainlessbot3699]

@@angelmendez-rivera351 I actually did know that.But thought of sharing the other one.

[@andriotik007]

9:07 didn’t get at all...

[@hOREP245]

The dominating power on the numerator is n^4, and the other powers will increase slower than n^4 (which is on the denominator). This means the only one that will matter is n^4.

[@98danielray]

the bigger the n, the more despiseable the other powers become next to n^4. if n tends to infinity, the tendency is for the bigger powers to dominate the value completely.

[@JensenPlaysMC]

If you dont get it, factor out the highest power on each of the terms. the terms with lower will be k/Infinity which equal zero meaning they dissappear

[@OtiumAbscondita]

First AGAIN

[@btdpro752]

For me you were the only comment with 2 likes

[@nestorv7627]

gay

[@spaghettiking653]

Subtitles: Japanese

[@JamalAhmadMalik]

in tea girl #yay

[@lumina_]

#yay

[@krutarthshah3302]

Can you give your email because I think I discovered something pretty neat. If you don't give, it's all right, I will write the latex in the comments

[@blackpenredpen]

MCandMore ByGamer Sure, it's [e-mail]

[@krutarthshah3302]

@@blackpenredpen Thanks!

[@krutarthshah3302]

@@blackpenredpen I sent you the mail!

[@danielgates7559]

second