HAVE YOU DONE THIS WITHOUT INTEGRATIONS BY PARTS?
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- เผยแพร่เมื่อ 4 มิ.ย. 2024
- This video discusses a popular integral, given as a definite integral for this example, which is usually evaluated using integrations by parts by many people. We will solve that problem using a new approach, which is applicable for definite integrals, with the lower limit of zero.
You can evaluate the indefinite integral with a matrix, since the span of xsinx, xcosx, sinx and cosx is closed under differentiation and integration. Super quick and can be applied to any linear combination of the functions, it's great.
Absolutely. It’ll be shown on a future video. Thanks for the idea. 👍🏻
Can you do a video about solving integrals with matrices, Polar coordinates, complex num?
@@mimorouidjali5487 Here's an example of how you'd solve an integral with a matrix.
Given:
integral (5*x^2 + 10*x + 7)/sqrt(x + 1) dx
Assume the solution has the form, called your Ansatz:
(A*x^2 + B*x + C)*sqrt(x + 1)
The reason we can assume this, is that each power increases by 1, and the square root term is maintained. As long as the solution isn't a logarithm or inverse trig, there will be an algebraic solution that is a product of a polynomial and a form of the original square root. If it involves a log or inverse trig, you'll get a degenerate matrix.
Take the derivative of the Ansatz:
1/(2*sqrt(x + 1)) * (A*x^2 + B*x + C) + (2*A*x + B)*sqrt(x + 1)
De-rationalize the denominator of the second term, to get it to look like the original integral:
1/(2*sqrt(x + 1)) * (2*A*x + B) + (A*x^2 + B*x + C)*sqrt(x + 1)
Expand and gather like terms:
(5/2*A*x^2 + (2*A + 3*B)*x + 2*B + C))/sqrt(x + 1)
Now we can match coefficients to the original integral:
5/2*A = 5
2*A + 3/2*B = 10
B + C/2 = 7
Which we can represent as a matrix equation:
[5/2, _ 0, _ 0] _ [A] _ [5]
[2, _ 3/2, _ 0] * [B] = [10]
[0, _ 2, _ 1/2] _ [C] _ [7]
And with our favorite matrix solving method, we can find the solution:
A = 2, B = 4, C = 6
Thus the solution is:
(2*x^2 + 4*x + 6)*sqrt(x + 1) + K
Great work, I love who you solve integrals without using the usual methods and we learn from it
Thank you so much for your wonderful comments....! I will try to make more videos.
Oh, The Famous Kings Property!
The famous Kings….! ✅
Great video, just subscribed after watching all your older videos too there are some excellent tricks here
I am glad you watched my videos. Thank you so much for your encouraging comment and subscription.
You could do the general case ...
Yes, a generalization of the theorem can be found here: CAN YOU EVALUATE THIS DEFINITE INTEGRAL?
th-cam.com/video/Kpstn9E1Ses/w-d-xo.html
That is a really clever trick, thank you. Can you please point me to the video where you show how the identity you used is derived?
Thank you so much. I think I proved the initial version here: th-cam.com/video/PYvHt7DF2cs/w-d-xo.html
@@gayansamarasekara Thank you. That also is clever. It occurs to me that for any definite integral, so long as neither limit of integration is infinity, then we can always make a substitution such that we have an integral from zero to something and then we can further apply this wonderful trick. It's mathemagical :)
@@simongross3122 Exactly....! Also, when we are a little too bored, we can fix the genes of the theorem, and have it ready for any asymmetric limits, such as the one discussed here: th-cam.com/video/Kpstn9E1Ses/w-d-xo.html, to be more mathemagical :) Thank you for your nice mathemagical comment....!
@@gayansamarasekara Haha my pleasure. I came across that particular phrase a long time ago when I worked in the IT industry. I'm not actually a mathematician although I have always had a keen interest, more in the philosophy than the practice. I'd rather be a mathemagician :)
@@gayansamarasekara I watched that video and I was not disappointed. I was expecting you to have a theorem that went Integral from limits a to b being replaced by integral limits from 0 to something (perhaps b-a), but you surprised me. :) I am pretty sure this also can be done.
Do this instead
Special case for such questions where we hsve xf(x) if f(x). Does not change after applying kings rule the. We can remove x by x=(upperlimit + lowerlimit)/2 and then Integra f(x)
Good Comment on Kings Rule. This video discusses an example from a popular form of problems found in calculus 1 classes, as a special case of the property, where the lower limit is zero.
Please make a video about yourself.
Thank you, I will keep that in mind. Basically, I am a professor at k-state, USA, I teach math and statistics classes and do research. I'm thinking to make videos for this channel to help the students in: college algebra, trig and calculus 1, 2, 3 courses taught for non-math undergrad majors of STE(M) fields of the US universities.