PokeVideoPlayer v23.9-app.js-020924_
0143ab93_videojs8_1563605_YT_2d24ba15 licensed under gpl3-or-later
Views : 11,806,310
Genre: Education
License: Standard YouTube License
Uploaded At Dec 28, 2022 ^^
warning: returnyoutubedislikes may not be accurate, this is just an estiment ehe :3
Rating : 4.867 (12,782/371,717 LTDR)
96.68% of the users lieked the video!!
3.32% of the users dislieked the video!!
User score: 95.02- Overwhelmingly Positive
RYD date created : 2024-11-21T22:19:54.77944Z
See in json
Top Comments of this video!! :3
The integral can be easily solved mentally. The term with the cosine function is point-symmetric with respect to the origin due to the factor x^3. Because of the symmetric limits of integration, the integral over this part vanishes. What remains is the factor 1/2 and an integral over a semicircle with a radius of 2. The area of the circle is 4Ο. Thus, the area of the semicircle is 2Ο. Due to the factor 1/2 in the integral, the result is Ο.
49 |
For those who did not understand, first he multiplied and expanded the given integral function into two integrals. For the first integral function (π₯^3)*cos(π₯/3)*sqrt(4βπ₯^2), he checked whether the function is odd or even. It is odd. If the function is odd from βπ to π, the value of the integral evaluates to 0. Hence, its integral became 0.
Now, for the second part (1/2)*sqrt(4βπ₯^2), we all know that definite integration is basically area under the curve.
He squared the function sqrt(4βπ₯^2) which resulted in the equation of a circle. He then calculated the area of the circle, but we only want the area under +sqrt(4βπ₯^2) and not above -sqrt(4βπ₯^2). Therefore, he took half of the area of the circle.
After further simplification, the final answer is π.
Edit: btw that ππ₯ should not be under the square root sign, so mathematically the questi
10 |
@martingu1005
1 year ago
This is exactly how math teachers expect us to use math "in the real world"
96K |