In this video, I showed how to integrtae the product of a polynomial and an exponential function
Пікірлер: 26
@knoxofficial2244 ай бұрын
NEW knowlodge unlocked✔
@Moj943 ай бұрын
the last one brings back some nightmares of mine.
@joelmacinnes23913 ай бұрын
Paused after 30 seconds, sped through the first 3 and spent ages contemplating the 4th, eventualy giving up and finding seconds later that I'm physically incapable of solving it
@CharashGanja3 ай бұрын
Exactly 😂
@prakrit12803 ай бұрын
Sir's smile 😇 proves his passion for mathematics👍😎👍Thank you for the informative video. Looking forward for your upcoming videos. 🙂
@RozheenBarwary2 ай бұрын
Good job 👏🏻👏🏻👏🏻👏🏻
@pratiksuryawanshi71884 ай бұрын
Amazing
@kevinduffy73833 ай бұрын
Would you (in time) be willing to talk about integration techniques for all non elementary integrals?
@paulor.r.correia17893 ай бұрын
Excelente 🇧🇷 🇧🇷 🇧🇷 🇧🇷
@TheExpertMathWithAnolog3 ай бұрын
1:46 OR use ilate which translate to inverse trig then logarithm, then algebra then trignomerty then the exponents
@kappascopezz51223 ай бұрын
1. Integration by parts: u=x, dv=e^x dx, du=dx, v=e^x. Therefore, I1 = int x e^x dx = x e^x - int e^x dx = (x - 1) e^x. 2. Substitution: Substitute u=x², and du=2x dx. Therefore, I2 = int x e^x² dx = 1/2 int e^u du = 1/2 e^u = 1/2 e^x². 3. Again the same substitution: u=x², du=2x dx which yields I3=int x³ e^x² dx = 1/2 int u e^u du. Applying the result from part 1 yields I3 = 1/2 (u-1) e^u = (x²-1)/2 e^x². 4. Again integration by parts: u=x, dv=xe^x² dx, du=dx, v=1/2 e^x² Therefore, I4 = int x²e^x² dx = x/2 e^x² - 1/2 int e^x² dx = x/2 e^x² - sqrt(pi)/4 erfi(x) Edit: Because the wikipedia page of the error function defines erfi as just erfi(z)=erf(i z)/i instead of erfi(z)=2/sqrt(pi) int e^x² dx here a short proof of the latter equality: erfi(z) = erf(i z)/i = 1/i int_0^(i z) e^(-x²) dx Now substitute x = i u, dx = i du = int_0^z e^u² du
@Player-pj9kt3 ай бұрын
Oh wow number 3 was surprisingly easy to solve
@darcash17383 ай бұрын
Could you do a video on greens theorem next? I heard of the term but I’ve been too lazy to find out what it is for a few days already. Maybe my excuse is that my finals are coming up in a few days but that’s still a bad excuse 😂
@davidmelville56753 ай бұрын
Greens theorem and Stokes theorem were two of my favourite things about that particular semester.
@darcash17383 ай бұрын
@@davidmelville5675Stokes also sounds pretty cool. I like it when they have cool names like the Wronskian. It makes em easier to remember bc when I randomly think of the name I also think of the concept and review it without realizing 😂
@joelmacinnes23913 ай бұрын
Ive never used or come across an error function, but is it used to compensate for the fact that the antiderivative of something like e^(x²) can't be solved by putting in e^(x²) × d/dx (x²) because d/dx (x²) isnt a product, and so ½xe^(x²) would give you a product when differentiated?
@golddddus3 ай бұрын
Never stop learning. I am now 74 years old. But I'm not so senile that I don't see that the solution of the fourth integral is completely wrong. Probably the minus in the exponent was overlooked in the task setting.😎
@PrimeNewtons3 ай бұрын
Thanks for the feedback, sir. What do you suggest as the correct answer?
@koshyboruto3 ай бұрын
Hi im french, ive been watching a lot of english maths videos and ive seen a lot of people talking about « calculus 2 » and things like this, how does the english studies works,( im not studying maths but i would like to learn theme like this and i dont know the equivalents in my country)
@PrimeNewtons3 ай бұрын
Calculus 1 starts from tangents, limits, and the power rule for differentiation. Then, it covers all techniques for differentiation. Covers applications of differentiation and introduces the student to integration. Fundamental Theorem of Calculus 1&2, Riemann sum, definition of the integral, u-substitution applications of integration. Calculus 2 covers the rest of single-variable calculus and all applicable integration techniques. Introduces power series, too.
@antonionavarro10003 ай бұрын
Me volví loco intentado resolver la última integral. Parece que son una pesadilla todas las integrales de la forma int { (x^m) • e^(x^2) } dx donde m es un entero par positivo. Mientras que si m es impar se calcula fácilmente mediante una fórmula de reducción, deducida mediante integración por partes. ¿Existe alguna manera de calcular la integral indefinida int { e^(x^2) } dx?
@apoorvgupta96803 ай бұрын
please it is possible for you to make videos on discrete mathematics or probability. because there are not much good content.
@adw1z3 ай бұрын
Last Integral: int[x^2 e^(x^2)] dx = int[ x * xe^(x^2) ] dx (Use parts) = 0.5xe^(x^2) - 0.5 int[e^(x^2)] dx int[e^(x^2)] dx = sqrt(pi)/2 erfi(x) + c, this is the definition of the imaginary error function erfi(x) == -i erf(ix); erfi(x) := 2/sqrt(pi) int(0 to x) [e^(x^2)] dx
@holyshit9223 ай бұрын
In Erfi i stands for imaginary
@omograbi3 ай бұрын
Even fast forwarding doesn't make the video fast enough
@Osirion163 ай бұрын
If you're not here for the explanations go straight for the answers, the guy is teaching for people to follow. Plug the integral in wolfram alpha if you just want the answer anyways