Integration By Partial Fractions (irreducible quadratic factors)
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- čas přidán 3. 07. 2024
- Dear calculus 2 students. Here we have 3 integrals of rational functions and they all have an irreducible quadratic factor in the denominator.
Blooper: • when calculus teacher ...
Here's how to set up the partial fraction decomposition: • Ultimate Partial Fract...
0:00 dear calc 2 students!
0:08 Q1 integral of x^4/(x^2+9)
3:15 Q2 integral of a big rational function
10:47 Q3 integral of a "bigger" rational function
Partial fraction decomposition playlist: bit.ly/calcPFD
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Bloopers: czcams.com/video/imiAEXDRDys/video.html
That is why calculators exist.
5:16, hmmmmm. I wonder was removed
Yeah idk what it could be, ah probably nothing major and embarassing
😂
And I love how he just emphasised *128* like he won’t let himself make a mistake this time
xD
I made the incredible mistake of reading “Calc 2” at the beginning as “Algebra 2”. Wow was I confused the entire video
i didnt make that mistake and im still confused LMAOO
@@locke8412LOL
In Romania we don't use those methods, but i love how you present it!! I had once a chance to use one of your methods in class and my classmates were shocked ;)) keep on the great work, professor!!!! Many good whishes from Romania!!
You make me love Calculus even more than I already do 😁
Last time i had to integrate something like these, i had an x^2+1 as factor in the denominator. I thought "ehh let's try using the cover-up method with complex numbers"'. Turns out it still works! The result will be a complex number in the from a+bi, but it's expectable, since you're looking for two coefficients in the numerator. So a and b will actually be B and C int the partial fraction.
My preferred method for his example at 3:20. Use strategic x-values, rather than setting up a system of equations, to find B & C.
(8*x^2 - 10*x - 3)/[(x - 4)*(x^2 + 1)]
Set up partial fractions, with the (x - 4) term first:
(8*x^2 - 10*x - 3)/[(x - 4)*(x^2 + 1)] = A/(x - 4) + (B*x + C)/(x^2 + 1)
Cover-up method for A, at x = +4:
A = (8*4^2 - 10*4 - 3)/[4^2 + 1] = (128 - 40 - 3)/17 = 5
Strategically let x = 0, and x = 1, to find B & C:
(8*x^2 - 10*x - 3)/[(x - 4)*(x^2 + 1)] = 5/(x - 4) + (B*x + C)/(x^2 + 1)
At x=0:
-3/[(-4)*(+1)] = 5/(-4) + C
3 = -5 + 4*C
C = 2
At x=1:
(8 - 10 - 3)/[(1 - 4)*(1 + 1)] = 5/(1 - 4) + (B + 2)/(1 + 1)
-5/(-3*2) = -5/3 + (B + 2)/2
5 = -10 + 3*B + 6
B = 3
Thus our partial fraction result is:
(8*x^2 - 10*x - 3)/[(x - 4)*(x^2 + 1)] = 5/(x - 4) + (3*x + 2)/(x^2 + 1)
Thanks.... was struggling with the same problem
Excellent solution ✌👍😀
I like it of how he holds a Pokemon mic
😂
Hello teacher
If you don't like integral and parenthesis next to each other, you can use bracket instead of parenthesis!
Or ...or you can use fatter integral (integral with obesity!)
I love you teacher
When will you upload that video
I love the poster in the background! Where can I get one??
5:11 If you're here from the bloopers video
I think for the first one trig sub works
x=3tan0
dx=3sec²0
tan⁴0/9sec²0 sec²0 d0
tan⁴0/9 d0
1/9 ×Integral(tan⁴0)
And then it gets absurd
(Integral of tan⁴0=tan³0/3 -tan0+x+C
=x³/3 +C
(I havent watched the video yet Im excited to see what he gets)
What about *as always: That's It* ❓❓
π'th
I wish you were my teacher ;)
How you use 2 pens at once
And switch
no IT guy in the house
Guys do any one has a prove of quadratic denominators have linear numerator? And on the same rythme
Using his example at 3:20:
(8*x^2 - 10*x - 3)/[(x - 4)*(x^2 + 1)]
Playing devil's advocate, assume that there is only a constant on top of the irreducible quadratic:
(8*x^2 - 10*x - 3)/[(x - 4)*(x^2 + 1)] = A/(x - 4) + B/(x^2 + 1)
Multiply to clear denominators:
8*x^2 - 10*x - 3 = A*(x^2 + 1) + B*(x - 4)
Expand:
8*x^2 - 10*x - 3 = A*x^2 + A + B*x - 4*B
Equate corresponding coefficients:
A = 8
B = -10
A - 4*B = -3
As you can see, we have 3 equations, and only 2 unknowns. A and B are both locked down by the first two equations, which means we can't satisfy the third equation. This is why we need another unknown, created by producing a linear numerator on top of the quadratic.
If we put a quadratic numerator on top of the quadratic, you'll end up with 4 unknowns, and 3 equations, which we won't be able to solve. Also, any time the degree on the top is equal to or greater than the degree on the bottom, we can always do long division to reduce it. Partial fractions requires terms that have at least one less degree on the top, than they have on the bottom, so they start reduced.
What’s man saying at 0:58
I had to turn on closed captions to figure that out lol, he's saying "and in fact we can continue"
That pokeball he is holding throughout the video
Why?
Just thinking of it and can't understand
Microphone you can see the wire
@@jackwang3006 yaa I have also one like that !!
Pokeball gives him the superpower to remember the formulae. 😉
Wrong Solution:
8*(4)²-10(4)-3= 144-40-3=101
Wrong. 8 * 16 = 128. he was right
@@vincenttran3342he was referring to a blooper video, where bprp said that 16 * 8 = 144 and was confused for 2 minutes
First
nice beard bro