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
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
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?
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.😎
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)
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.
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?
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 😂
@@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 😂
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
NEW knowlodge unlocked✔
the last one brings back some nightmares of mine.
Sir's smile 😇 proves his passion for mathematics👍😎👍Thank you for the informative video. Looking forward for your upcoming videos. 🙂
Great content from a great teacher.
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
Exactly 😂
Good job 👏🏻👏🏻👏🏻👏🏻
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
Would you (in time) be willing to talk about integration techniques for all non elementary integrals?
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?
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.😎
Thanks for the feedback, sir. What do you suggest as the correct answer?
1:46 OR use ilate which translate to inverse trig then logarithm, then algebra then trignomerty then the exponents
Oh wow number 3 was surprisingly easy to solve
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)
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.
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?
Excelente 🇧🇷 🇧🇷 🇧🇷 🇧🇷
please it is possible for you to make videos on discrete mathematics or probability. because there are not much good content.
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 😂
Greens theorem and Stokes theorem were two of my favourite things about that particular semester.
@@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 😂
Amazing
In Erfi i stands for imaginary
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
Even fast forwarding doesn't make the video fast enough
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