In question 3, does we know the second one is equal to positive infinity? Therefore negative infinity and positive infinity will cancel each other. So we get a finite number and it is convergence.
6:41 in case v.a. gained is not equivalent to any upper or lower limit then break the integral into two improper integral with gained one being the middle one
If the function is not continuous at the upper limit b, then t approaches b from the left since the interval is [a, b), which are values less than b. Similar idea for the lower limit.
But how can you tell where exactly the limit is continuous or discontinuous between the upper limit and the lower limit? Or maybe you can choose between the two which one to make discontinuous and continuous, if so then can you find the same answers
@@PurityMutinta-zd4reyou just need to look at the problem, wherever plugging a certain number makes the whole integral undefined that’s what you’re gonna use for a limit
We chose 4 because the function is not continuous at 4; and we don't choose 0 because the function is continuous at 0. For the function for example 2, 1/cbrt(4-x), we can see that the function is undefined at 4 since the denominator becomes 0 when we substitute 4.
@@lperezherrera1608 The earlier comments were not talking about the same thing - The integrand was the same while the intervals of integration were not.
Phenomal lesson, I have a test next period and you just made my day. Thanks a ton! Very clear and concise.
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WOW! you are a great teacher. so clear. You need way more subscribers
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thanks for the clear practice!
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Hey! Tysm actually this video is really easy to understand thanks again😀
Good to know! Happy New Year!! 😁
In question 3, does we know the second one is equal to positive infinity? Therefore negative infinity and positive infinity will cancel each other. So we get a finite number and it is convergence.
Infinities are not numbers. We can't "cancel" them.
Excellent video. Very well organized and explained for the 3 examples of type II improper integrals. Thank you!
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Thanks a lot sir... U r an angel....😇.....
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Great videos very clear and concise explanations 👍🏿
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Thank you so much, you're a life saver ❤❤❤❤
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6:41 in case v.a. gained is not equivalent to any upper or lower limit then break the integral into two improper integral with gained one being the middle one
Yeah
thank you!!
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ilysm
How do we know it limit is approaching from right side or left skde
If the function is not continuous at the upper limit b, then t approaches b from the left since the interval is [a, b), which are values less than b. Similar idea for the lower limit.
But how can you tell where exactly the limit is continuous or discontinuous between the upper limit and the lower limit? Or maybe you can choose between the two which one to make discontinuous and continuous, if so then can you find the same answers
@@PurityMutinta-zd4reyou just need to look at the problem, wherever plugging a certain number makes the whole integral undefined that’s what you’re gonna use for a limit
Thank you so much, I have one question - how do you know which one is discontinuous, like in example -2 why we didn't choose 0 instead of 4 ?
We chose 4 because the function is not continuous at 4; and we don't choose 0 because the function is continuous at 0. For the function for example 2, 1/cbrt(4-x), we can see that the function is undefined at 4 since the denominator becomes 0 when we substitute 4.
How do you know it’s approaching negative infinity, I don’t get it ?
It's given in the integral, that's how we know.
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Question, in the last one, one of them only diverges, doesn’t that mean we should see if the second once also diverges or converges?
No if one of them diverges the integral diverges
As what Riley said 😁👍
@@lperezherrera1608 The earlier comments were not talking about the same thing - The integrand was the same while the intervals of integration were not.
my proffessor sucks in my calc 2 class i can barely understand bro
I hope you are getting better at this!
ilysm