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Leaving Cert 2016/17 Off-Topic Thread
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No way? That's disappointing to hear. Last year two of our teachers didn't come to graduation. One had a genuine excused and apologised to us that she wouldn't be able to attend. The other just didn't care. Says a lot really about her really. Graduation was honestly the best night out I've ever had. Having a drink with your school friends and teachers is a surreal experience and I honestly enjoyed it thoroughlyHad my graduation yesterday, felt kinda underwhelming with some teachers failing to show up and a feeling of a general lack of planning associated with the event.
Anyone else feel the same?
Now that the LC's so soon things are really starting to heat up though.
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We did tell the principal and the yearhead, the announced the ceremony had ended without giving us a chance to do it. As for prefects, well they are mostly just prefects in name only.
Thanks for the suggestion, but I amn't going to debs and just glad its over now. Its just something that irritated me.
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X6.430macman wrote: »My last day was the worst ever..no one did anything and it was so boring and bad
Sorry to hear that, I hope you have a good summer holiday to compensate!
I know the feeling of being dissatisfied but I take a zen approach to it
Does anyone know where we can find sample questions and stuff for the supplementary calculus or is it in the booklet, cause idk where mine is.
Also what are people's predictions for English? Especially the single text (Hamlet) and the poets (my bet is Hopkins or Plath)0 -
Sorry to hear that, I hope you have a good summer holiday to compensate!
I know the feeling of being dissatisfied but I take a zen approach to it
Does anyone know where we can find sample questions and stuff for the supplementary calculus or is it in the booklet, cause idk where mine is.
Also what are people's predictions for English? Especially the single text (Hamlet) and the poets (my bet is Hopkins or Plath)
I think Hopkins or Donne. Possibly Boland too.0 -
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Take Your Pants Off wrote: »Anyone know if integration by parts is still on maths course. Because i was looking ay b part i. I presume thats hoe you go about doing the qs. But my friend said its no longer on the course.
Also 8 part a. I dont quite get how the answer is 2ln(x+3) maybe using the second rule in integration in formula booklet ?
You can do (b) (i) using a trigonometric substitution on the Formulae and Tables Booklet. It's the second one from the top on page 15. I've attached a solution to that one. You can also use substitution, but for some crazy reason, it's not on the course. It's unfortunate since it's actually quite helpful and doesn't require too much learning. Please let me know if any of it is unclear and I'll try to explain.
I don't know how to do (a) (i) without applying some form of substitution. I'll try to explain it conceptionally and simply.
The function is 2 / (x + 3). 3 is just a constant. So you could simply let u = x + 3. Now you have 2 / u, which you know integrates to 2ln|u| + C. But since u = x + 3, you can say that the integral is 2ln|x + 3| + C.
Here is a bit more information in the spoiler which is more than you need to do the question but might make some things a bit clearer. If any of this doesn't make sense, please just ignore it. It's here to make you understand a bit better, but it's not essential for Leaving Cert Maths. Just highlight the text with your cursor to reveal the text.Essentially all you need to know is that something like x + a, where a is a constant like 3, pi, or -6.25, can be written as some other function u. So for x + 3, if x = 1, it evaluates to 4. That means that u = 4. The concept might be a bit confusing, giving x + 3 the value u tidies things up a bit for you.
In general, the indefinite integral of a / (x + b), where a and b are constants, equals aln|x + b| + C.
I don't think you need to know this, but strictly speaking, the integral of 1 / x is ln|x|, with absolute value or modulus bars. This can be proved, but essentially it accounts for when x is a negative value. Since you can't calculate the log of a negative value, the absolute value bars ensure that whatever number x is, is a positive number, excluding 0 of course.
Also, strictly speaking, what I told you above would give you the indefinite integral of 2 / u dx. You can't integrate this, but you can differentiate u with respect to x. In this case, du/dx of x + 3 is just 1. So you can basically multiply across by dx to get the equation du = dx. So you can change your integral to 2 / u du, which you can integrate.
Remember that u is some other variable. There is nothing special about x. It's just what mathematicians, scientists and engineers frequently use. You can differentiate and integrate with respect to any variable.
This website is very useful for calculus solutions, but bear in mind that it does use substitution, integration by parts and a bunch of stuff you don't know in some of its solutions.
https://www.symbolab.com/solver/step-by-step/\int\frac{2}{x+3}dx
Don't forget the constant of integration in indefinite integrals!0 -
You can do (b) (i) using a trigonometric substitution on the Formulae and Tables Booklet. It's the second one from the top on page 15. I've attached a solution to that one. You can also use substitution, but for some crazy reason, it's not on the course. It's unfortunate since it's actually quite helpful and doesn't require too much learning. Please let me know if any of it is unclear and I'll try to explain.
I don't know how to do (a) (i) without applying some form of substitution. I'll try to explain it conceptionally and simply.
The function is 2 / (x + 3). 3 is just a constant. So you could simply let u = x + 3. Now you have 2 / u, which you know integrates to 2ln|u| + C. But since u = x + 3, you can say that the integral is 2ln|x + 3| + C.
Here is a bit more information in the spoiler which is more than you need to do the question but might make some things a bit clearer. If any of this doesn't make sense, please just ignore it. It's here to make you understand a bit better, but it's not essential for Leaving Cert Maths. Just highlight the text with your cursor to reveal the text.Essentially all you need to know is that something like x + a, where a is a constant like 3, pi, or -6.25, can be written as some other function u. So for x + 3, if x = 1, it evaluates to 4. That means that u = 4. The concept might be a bit confusing, giving x + 3 the value u tidies things up a bit for you.
In general, the indefinite integral of a / (x + b), where a and b are constants, equals aln|x + b| + C.
I don't think you need to know this, but strictly speaking, the integral of 1 / x is ln|x|, with absolute value or modulus bars. This can be proved, but essentially it accounts for when x is a negative value. Since you can't calculate the log of a negative value, the absolute value bars ensure that whatever number x is, is a positive number, excluding 0 of course.
Also, strictly speaking, what I told you above would give you the indefinite integral of 2 / u dx. You can't integrate this, but you can differentiate u with respect to x. In this case, du/dx of x + 3 is just 1. So you can basically multiply across by dx to get the equation du = dx. So you can change your integral to 2 / u du, which you can integrate.
Remember that u is some other variable. There is nothing special about x. It's just what mathematicians, scientists and engineers frequently use. You can differentiate and integrate with respect to any variable.
This website is very useful for calculus solutions, but bear in mind that it does use substitution, integration by parts and a bunch of stuff you don't know in some of its solutions.
https://www.symbolab.com/solver/step-by-step/\int\frac{2}{x+3}dx
Don't forget the constant of integration in indefinite integrals!
Thank you so much for taking your time out and explaining.0 -
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