Integration Maths Assignment Question - Page 1
Forum section: electronics / beginners

author: Stewboi
author_rank: Newbie
timestamp: July 06, 2015, 09:35:57 pm
content: Hi, I am in the first year of my higher national certificate in Electrical Engineering at college. This is a question from my integration assignment and is the final assignment of this maths unit. I have attempted the question as best I can but believe I am going wrong somewhere with my calculations because when I attempt to do part 3b I cannot get an answer. All and any advice is welcome.I have uploaded 3 attachments - they are of the question I am stuck on, the attempt at the first and second part of question 3a.Any advice is welcome. If you need the images to be bigger I can upload bigger ones but I decided to upload them slightly smaller to save on the size of the attachments.Thank you in advance,Stewart

author: Whales
author_rank: Super Contributor
timestamp: July 06, 2015, 10:50:25 pm
content: Does your course cover trigonometric substitutions for the denominators?EDIT: No, I think I now recall that this is only useful when you substitute inside a square root.

author: Andy Watson
author_rank: Super Contributor
timestamp: July 06, 2015, 10:58:08 pm
content: Watch thebirdyintegral sign.

author: Andy Watson
author_rank: Super Contributor
timestamp: July 06, 2015, 11:03:00 pm
content: Quote from: Whales on July 06, 2015, 10:50:25 pmI was a bit phased by that. I think what they mean by substitution is taking 1/3 outside of the (3x-2). i.e. (3x-2) is substituted with (x-2/3) and scaled by 1/3. Then the logarithmic integration can be applied again.

author: IanB
author_rank: Super Contributor
timestamp: July 06, 2015, 11:29:40 pm
content: I think the substitution u = 3x - 2 works fine for integrating 1/(3x-2). You get du = 3 dx and then you can integrate 1/3 (1/u) du. It looks like you got it right, but as Andy said you forgot to take away the integral sign after you had integrated.

author: Howardlong
author_rank: Super Contributor
timestamp: July 06, 2015, 11:47:47 pm
content: I hope the OP is able to get some practical time in too!Why do academics insist on making up these contrived math(s) examples? A practical example would be of so much better value.Grrrr, still fretting 30 years after my EE degree, man I could do calculus 'til the cows come home back then, but I never once got to use a scope in a lab.Now between calculus and a scope, which do I use most in the real world on a 1000:1 ratio?

author: tec5c
author_rank: Frequent Contributor
timestamp: July 07, 2015, 12:03:08 am
content: I can confirm your working is correct, but as mentioned you have left the integral sign in your (working/)solution. They, the marker, might be a bit forgiving in an exam but as this is an assignment, I would assume you would be deducted marks for this. At least, this is how it worked for me when I did math assignments.As for part b, you use the same substitution but as you're now doing a definite integral you need to be cautious of what the integral limits will become once you do the substitution. You also need to do a second sub. for the integrand 1/(x+1) which will also give different upper and lower limits (differing from the given -0.5 to +0.5).My best advice is to just be careful with your upper and lower limits when doing U-substitution. Otherwise, the integration is trivial.

author: IanB
author_rank: Super Contributor
timestamp: July 07, 2015, 02:09:43 am
content: Quote from: tec5c on July 07, 2015, 12:03:08 amI would suggest reversing the substitution after doing the integration in order to avoid getting the numbers in a tangle.

author: IanB
author_rank: Super Contributor
timestamp: July 07, 2015, 02:18:31 am
content: Quote from: Howardlong on July 06, 2015, 11:47:47 pmI was thinking over the same thing. However, elements of calculus like this are single pieces in a huge web of knowledge. Each piece by itself may seem to have no more significance that a single brick in a wall, but take away too many bricks and the whole building comes tumbling down. The more pieces you have in your web of knowledge, the easier it is to absorb and understand new concepts as you continue learning. So the time spent doing such exercises is not wasted.Also, if you can do calculus 'til the cows come home you should have no trouble at all learning how to use a scope. From never having touched a scope in my life, it only took me a few hours to get the basic controls and settings sorted out. Learning what the knobs and menus were for was made much easier by having a theoretical understanding of what is going on with AC and DC signals, frequency, amplitude, bias and so on.

author: Simon
author_rank: Unknown
timestamp: July 07, 2015, 05:46:00 am
content: Quote from: Howardlong on July 06, 2015, 11:47:47 pmI don't know where the OP is studying but I'm just being enrolled on my HNC having done one foundation module in maths that I've mostly forgotten months later. The HNC course has 3 modules called "analytical methods for engineers" aka more bloody maths with a fancy name, then just 4 in electronics and a module for a project that has me puzzled as the electronics modules are all optional choices so I don't know how they will chose the project to suit any combination of 4.

author: Howardlong
author_rank: Super Contributor
timestamp: July 07, 2015, 07:11:58 am
content: Quote from: IanB on July 07, 2015, 02:18:31 amWell I was lucky enough to have occasional access to scopes when I was a nipper, before I ever went to university, through the kindness of one of my older school buddy's connections. I was about twelve or thirteen when I first probed my projects which at that time were already forty or fifty hand wired chips worth. Regrettably for others on my EE course, they were not so lucky, most of them made it to the mortar board without seeing either a scope or a soldering iron!But yes, I understand the building block concept, it's just that after my own academic experiences I often question whether the balance is right.

author: Howardlong
author_rank: Super Contributor
timestamp: July 07, 2015, 07:17:17 am
content: Quote from: Simon on July 07, 2015, 05:46:00 amThere's a nice quote on the Wikipedia page for Analytical Skills:Quote

author: tec5c
author_rank: Frequent Contributor
timestamp: July 07, 2015, 10:23:21 pm
content: From my second year onwards in my degree, I was constantly exposed to using an oscilloscope and also a spectrum analyser for the practical classes related to certain subjects. Along with simulating a circuit, then building on a breadboard, then finally soldering the circuit on a piece of veroboard.I'm not sure if this is now common for University's in Australia or if it's just the Uni I am at.Mind you, being exposed to equipment is different than being taught how to use it. A lot of students struggled to use the gear correctly (spectrum analyser was a big stump for a lot of students)

author: Bob F.
author_rank: Regular Contributor
timestamp: July 08, 2015, 01:01:43 pm
content: Quote from: Howardlong on July 06, 2015, 11:47:47 pmLol! I recall sitting through an hour lecture a similar time ago while the lecturer derived Ohm's law from Maxwell's equations. To this day I have no idea why he did that, other than to show that he could. Such feats are no doubt fun for mathematicians, but it's wasted on the rest of us when we have real things we need to get on with. To be fair, we had at least one half-day lab session per week - plus project work so this was probably just his party-piece.

author: ScoobyDoo2
author_rank: Newbie
timestamp: July 08, 2015, 01:55:04 pm
content: The x=-0.5 limit makes no sense because the ln(3x-2) term cannot be evaluated.

author: Andy Watson
author_rank: Super Contributor
timestamp: July 08, 2015, 02:23:26 pm
content: Quote from: ScoobyDoo2 on July 08, 2015, 01:55:04 pmThere are two routes around that problem: Either follow the integration through to the end algebraically - at which point the negatives cancel out, or evaluate the negative log with complex arithmetic.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: July 08, 2015, 09:48:56 pm
content: Quote from: IanB on July 06, 2015, 11:29:40 pmint ((1 . dx)/ (3x-2)u = (3x-2)du = 3-2 dx = 1* dxint( du/u) = ln(u) = ln(3x-2)JP

author: Andy Watson
author_rank: Super Contributor
timestamp: July 08, 2015, 10:03:12 pm
content: Quote from: J4e8a16n on July 08, 2015, 09:48:56 pmThis forum software is useless for expressing maths, but I think that should be du = 3 dx. Otherwise you will be a factor of 3 wrong in the answer.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: July 09, 2015, 01:06:26 am
content: Sorry,3x -23 - 0

author: CatalinaWOW
author_rank: Super Contributor
timestamp: July 09, 2015, 01:20:13 am
content: Quote from: Howardlong on July 06, 2015, 11:47:47 pmWhile the ratio of scope time to calculus time was not that different for me in my career, the few times a year I would put the calculus to use was of such great value that it evened the balance substantially. I (and others) were often able to solve problems on the job that had stumped people who had left their maths back at the university. There seemed to be a strong correlation between this ability and annual compensation.The same kind of relationship continues to apply in this day of simulations for everything from analog circuits, to mechanical designs and even aerodynamics. Those who merely know how to enter the data and push the button are handicapped relative to those who know the inner workings of how the answer came to be.

author: rs20
author_rank: Super Contributor
timestamp: July 09, 2015, 01:25:01 am
content: Quote from: CatalinaWOW on July 09, 2015, 01:20:13 am+1000.