I was hoping this was going to be shown in the "long way", however the first part of splitting the equation into 2 individual equations with individual bounds helped answer my question and got you a like.
Sir, I am very thankful for the incredibly clear and logical explanations of the concepts. So many instructors prefer a muscle memory approach. This method makes it easier to apply the information.
5:54 why are you including in the second integral the value 3, I thought that X needs to be less than three not equal? Thanks. Mayb form -1 to 3+delta and then limit as delta goes to zero? Or else?
Nice video brother. But the last 2 maths you showed were wrong. Because the functions are not even continuous over the concerned intervals. So they are not integrable and the fundamental theorem of calculus doesn’t apply here. Other than that, your handwriting on the blackboard was very satisfying and the method of evaluating the integrals you showed is gonna help me a lot. Thanks.
there is 2 mistakes done in equation 2 and 3: in equ 2: x cannot be 0, so the condition should be x>0 in equ 3: x cannnot be 1, so the condition should be x>1
I am curious how you got that x can be greater than or equal to 1, in the last example? Doesn't 1 cause a vertical asymptote at x equal to 1? Why didn't you take the limit as x approaches 1 from both sides?
Hi Professor !!!! I have a function with absolute value how can you please help me the following: f(x) = e^|x| + lnx I am asked to calculate . The direction of variation - infinity to + infinity . Definition domain . Calculate the straight line of equation . The equation of the tangent . The area of the function . Graphical representation
I have a question. In the second exercise, we have |x|(x^2+1)/x; and you mentioned that this expression is positive when x > or equal than 0, but what if x=0? Then we will have an indetermination of 0/0. I think that that expression is positive only when x>0 and not equal. Is this right?
Thanks a lot for your video, my exam is about to start in 5 hours. However, for example 3, since F(x)=(x-1)(x+2)/|x-1| is not continuous in the close interval (-2,4) because of x≠1, does the integral not exist?
What would you do if you had the absolute value of x times cos(nx) dx ? I'm not sure how you would seperate this and make 2 integrals of it. But anyways great video enjoyed it very much!
@@PrimeNewtons Damn I guess my calc teacher is giving us hard stuff on purpose because we are expected to find the integral from 0 to pi/2 of |8 sin(x) − 8 cos(2x)| dx
The second and third integrals are improper integrals, because the integrands are not defined over the domains of integration. However, in these cases the integrands are only discontinuous on a point, which means that their integrals can still be evaluated by taking the limits of the integrands at their respective points of discontinuity. For example, since 𝑓(𝑥) = 𝑥 ∕ 𝑥 is not defined for 𝑥 = 0, ∫[−1, 1] (𝑥 ∕ 𝑥)𝑑𝑥 is to be understood as lim 𝑎→0⁻ ∫[−1, 𝑎] (𝑥 ∕ 𝑥)𝑑𝑥 + lim 𝑏→0⁺ ∫[𝑏, 1] (𝑥 ∕ 𝑥)𝑑𝑥. I understand that this was not the topic of this video, but I still find it important enough not to gloss over.
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Thank you for such a thorough, clear explanation. I am taking AP Calculus BC as a rising Junior and this was very helpful. Have a great day!
Good luck!
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Hello there!
Your videos have helped me gain confidence in calculus
I was hoping this was going to be shown in the "long way", however the first part of splitting the equation into 2 individual equations with individual bounds helped answer my question and got you a like.
Sir, I am very thankful for the incredibly clear and logical explanations of the concepts. So many instructors prefer a muscle memory approach. This method makes it easier to apply the information.
I LOVE YOUR HAT. THIS VIDEO WAS SO AWESOME AND EDUCATIONAL. THANK YOU
Thanks
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Subscribed + liked because u really deserve it!!!!! great video sir
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Gracias por el vídeo. Me gustó. Saludos desde Perú
Thank you very much for your instructive video ❤
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5:54 why are you including in the second integral the value 3, I thought that X needs to be less than three not equal? Thanks. Mayb form -1 to 3+delta and then limit as delta goes to zero? Or else?
Another excellent video! 😃
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Thank you!
Nice video brother. But the last 2 maths you showed were wrong. Because the functions are not even continuous over the concerned intervals. So they are not integrable and the fundamental theorem of calculus doesn’t apply here. Other than that, your handwriting on the blackboard was very satisfying and the method of evaluating the integrals you showed is gonna help me a lot. Thanks.
there is 2 mistakes done in equation 2 and 3:
in equ 2: x cannot be 0, so the condition should be x>0
in equ 3: x cannnot be 1, so the condition should be x>1
You are correct. I should have excluded 0 and 1 respectively
@@PrimeNewtons i love this video, it came in the right time for me, because i have math test next month, thank you
great video and nice hat
thanks uncle 👍🏾
I am curious how you got that x can be greater than or equal to 1, in the last example? Doesn't 1 cause a vertical asymptote at x equal to 1? Why didn't you take the limit as x approaches 1 from both sides?
extraordinary
i had no idea thank you
Hi Professor !!!!
I have a function with absolute value how can you please help me the following: f(x) = e^|x| + lnx
I am asked to calculate
. The direction of variation - infinity to + infinity
. Definition domain
. Calculate the straight line of equation
. The equation of the tangent
. The area of the function
. Graphical representation
10/10 for explicat°
I have a question. In the second exercise, we have |x|(x^2+1)/x; and you mentioned that this expression is positive when x > or equal than 0, but what if x=0? Then we will have an indetermination of 0/0. I think that that expression is positive only when x>0 and not equal. Is this right?
You are correct. That was a criminal oversight
Those who stop learning have stopped living I like that 🤔😁
Very well explained, thank you for this sir!
can you do (x^2)*(cos(3x)), from 0 to 2 pi?
it will take you long, so im wondering if you can do it quicker
Can you please re-do the equations with the mistakes or refer me to similar videos of yours to show how to correct them?
Thanks a lot for your video, my exam is about to start in 5 hours. However, for example 3, since F(x)=(x-1)(x+2)/|x-1| is not continuous in the close interval (-2,4) because of x≠1, does the integral not exist?
Cancel out the (x-1) . The integral exists. That's a removable discontinuity
for some reason i got 50/3 on example 2?
Try going over each step of your work again. I'm sure you'll find your mistake. You should get -22/3
Could you help me calculus?
💓💪
Thank you!
to the point thanks
🥰🥰🥰
Thanks
What would you do if you had
the absolute value of x times cos(nx) dx ?
I'm not sure how you would seperate this and make 2 integrals of it.
But anyways great video enjoyed it very much!
You will get that in advanced calculus. So, as far as calculus 2 is concerned, this video is sufficient.
@@PrimeNewtons Damn I guess my calc teacher is giving us hard stuff on purpose because we are expected to find the integral from 0 to pi/2 of |8 sin(x) − 8 cos(2x)| dx
this is cool
funny how you can learn more from TH-cam than in class, lol.
😊
Wow genial
There are so many ads in you videos, jeez
Chimwela
The second and third integrals are improper integrals, because the integrands are not defined over the domains of integration.
However, in these cases the integrands are only discontinuous on a point, which means that their integrals can still be evaluated by taking the limits of the integrands at their respective points of discontinuity.
For example, since 𝑓(𝑥) = 𝑥 ∕ 𝑥 is not defined for 𝑥 = 0,
∫[−1, 1] (𝑥 ∕ 𝑥)𝑑𝑥 is to be understood as
lim 𝑎→0⁻ ∫[−1, 𝑎] (𝑥 ∕ 𝑥)𝑑𝑥 + lim 𝑏→0⁺ ∫[𝑏, 1] (𝑥 ∕ 𝑥)𝑑𝑥.
I understand that this was not the topic of this video, but I still find it important enough not to gloss over.