Andrew 5 Report post Posted April 19, 2016 (edited) Hi, Mi Maths HL book (Pearson's Baccalaureate) contains a multivariate calculus question in the options section for calculus, e.g. the P3 section, and I'm just taking horrible amounts of time to finally solve it and I wonder if my P3 is going to require me to solve things of this kind. The question was: the definite integral from 0 to k of (x/(sqrt(k^2 - x^2)) I have had to do partial fractions, several layers of integrations by parts, trigonometric substitutions, weierstrass substitutions, and uncountable other substitutions, and this exercise alone would easily have taken me more than 2 hours. I'm dying 🙉 Edited April 19, 2016 by Andrew Share this post Link to post Share on other sites

bb8-m8 76 Report post Posted April 19, 2016 could you post a picture of the question? Share this post Link to post Share on other sites

kw0573 1,241 Report post Posted April 19, 2016 The question is not a multivariable calculus question. As far as the question goes, k is a constant. IB often uses letter (symbolic) representation for constants so it's not fair to call every question with more than one symbol representations a multivariable question. You simply had to do u = k^{2}-x^{2} and du = -2x dx. The key part is recognize the bound changes, which does come up time to time in P1 and P2. Put simply, you substitute the bounds as x in the expression for u to create two new bound expressions. It's certainly not unheard of to spend 2 hours on a single question in homework and I applaud your persistence. Hopefully through doing this problem have consolidated your knowledge and mastery of u substitution (which is in core HL syllabus), so when you do see a similar question on the exam you can be more prepared. For this particular problem multivariable calculus may have worked, but u substitution is sufficient and accessible. Share this post Link to post Share on other sites

Emilia1320 140 Report post Posted April 20, 2016 K is a constant, not variable. Only variable there is X. K is arbitrary positive real number. Share this post Link to post Share on other sites

Andrew 5 Report post Posted April 20, 2016 Ok. That k is constant does change it a lot. I had integrated by parts and then done the partial fraction and then the trig substitution just to get a thing i have to integrate by parts again and the the same thing. Then i gave up Share this post Link to post Share on other sites

kw0573 1,241 Report post Posted April 20, 2016 @Andrewscroll up to 3rd comment see my response. Let u = k^{2}-x^{2}, du = -2x dx should do the trick 1 Share this post Link to post Share on other sites

Andrew 5 Report post Posted April 20, 2016 3 hours ago, kw0573 said: @Andrewscroll up to 3rd comment see my response. Let u = k^{2}-x^{2}, du = -2x dx should do the trick thank you. It indeed does. Share this post Link to post Share on other sites