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Math HL

how do you integrate:

x arcsin(x)

with respect to x?

∫ x arcsin(x) dx = ??

I tried integrating by parts once and then by substitution twice but ended up with a very weird answer. I wonder if anybody knows how to integrate it like once or twice only instead of thrice. it would be great if they could also write it on a piece of paper and scan it because I guess the solution might be a bit long.. thank you!

My loooong solution:

longo.png

At some point, I used something our teacher called "the neat trick":

a = b + c - a

2a = b + c

a = 1/2*(b+c)

Wolfram|Alpha's method: http://www.wolframalpha.com/input/?i=integrate+x+arcsin+x

Edited by Gene-Peer
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Pre-IB student preparing to take Higher Level maths (next year) here.

While I generally enjoy mathematics and can easily pick up new concepts, I find myself frequently struggling with questions where I have to show / prove that a = b or some other equation. My biggest problem is figuring out where to start, since it's difficult for me to organize all the numbers that they give you into a coherent equation (or formula) with which I can prove the mathematical statement presented. Is there some way that I can practice these such questions, or a resource that I can use to help me with these types of problems?

Well, mathquestion.jpg

here's an example of a question that I'm having trouble with.

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You should look at them as "Find a". Ignore the other value for the moment. Then when you're done, you might find yourself with a=b, if not and you've made no mistake, then you'll have a=c where c can be simplified to b. If you can't simplify c, improvise by trying (on a rough paper) to rewrite b to c, and then on your paper just copy the steps in reverse. If you still can't do it, then just write it like:

a=c

=b

At most you'll miss two marks for working, but usually you might get away with it as the examiner might think you skipped the steps due to time constraints. :(

for the question:

a)

We know that the circle has radius x

So the length of the string used for the circle = circumference of the circle = 2pi*x

There for the perimeter of the rectangle = 32 - 2pi*x

The two "widths" of the rectangle have a combined size/length of 8

Therefore the two "lengths" of the rectangle have a combined size of 32 - 2pi*x - 8 = 24 - 2pi*x

So "one length" = 12 - pi*x

b)

Area of rectangle = length*width = (12 - pi*x)4 = 48 - 4pi*x

Area of circle = pi*x2

Total area = 48 - 4pi*x + pi*x2

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Thanks for helping me with that question! My problem was that I didn't know where to start; the first thought that came to my mind was "So what if the radius of that circle is x? How does that relate to anything?" I never thought that x can be used to find the circumference, and that added to the perimeter of the rectangle would equal the original length of the wire.

I think I need more practice with these types of "show a = b" questions.

A lot of people at my school say that Higher Level maths is difficult and many of them score very low on tests. How challenging is this course? What are the types of questions that are asked on tests?

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Thanks for helping me with that question! My problem was that I didn't know where to start; the first thought that came to my mind was "So what if the radius of that circle is x? How does that relate to anything?" I never thought that x can be used to find the circumference, and that added to the perimeter of the rectangle would equal the original length of the wire.

I think I need more practice with these types of "show a = b" questions.

A lot of people at my school say that Higher Level maths is difficult and many of them score very low on tests. How challenging is this course? What are the types of questions that are asked on tests?

From experience, HL maths is something you will find quite manageable or quite hard, and it just depends on the individual. Saying that, don't think 95% of people struggle with it. Very few that are actually capable to do it do struggle. The problem is it seems all the horror stories seem to stick while anything positive about HL maths seems to be forgotten. About the type of questions, well the best way to know is simply just do past paper after past paper from your teacher to get an idea. Its just like any subject, if you are familiar with the formatting of questions, they are going to make a whole lot more sense.

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help!!

if you have a quadratic expression, like let's say y=ax²+bx+c, where a,b,c are constants, is there any way to express x in terms of y?

my problem is I have an equation:

x²+(y+r+t)²=t² where r and t are constants

I need to express y² in terms of x, r and t, because I need to integrate y² with respect to x.

help, anybody? or is that even possible to be done?

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Well, you can rewrite it as 0 = ax2 + bx + (c-y) and use the quadratic formula.

x = 1/2a*(-b +- sqrt(b2 - 4a(c-y)))

As for your question, it is easier to make y subject of the formula:

x2 + (y+r+t)2 = t2

(y+r+t)2 = t2 - x2

y+r+t = +- sqrt(t2 - x2)

y = sqrt(t2 - x2) - r - t

or

y = -sqrt(t2 - x2) - r - t

Assuming you want to find that integral for the volume of revolution about the x-axis, the function you have is a circle with center (0,-(r+t)) and radius t. If you rotate it about the x-axis you form a shape that is similar to folding a cylinder of length 2pi(r+t) and radius t. This shape is known as a torus (or a doughnut, lol).

The volume of this cylinder = base area*length = pi*t2*2pi(r+t) or 2pi2t2(r+t)

This will work only if r + t > t, otherwise, you won't form a torus. This means r has to be positive.

You might want to look into Pappus' Centroid Theorem

Edited by Gene-Peer
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ohhhhh haha thank you! I didn't see that :) I expanded (y+r+t)² and was stuck :P

yup, the original question was: find the formula of the volume of a doughnut. teacher wants us to use integration :D

2t=the thickness of the doughnut

r=the radius of the hole of the doughnut

ok now since

y = ±√(t²-x²) - (r+t)

doesn't that mean that

y² = t² - x² ±2(r+t) + (r+t)²

?

what should I do to the ± sign? can I just take its positive value and then integrate it or just leave it and integrate it as it is?

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Well, it's actually y2 = t2 - x2 ±2(r+t)sqrt(t2-x2) + (r+t)2

The plus-minus sign means you have two curves, i.e. the top half of the circle and the bottom half of the circle. To find the volume between the two curves you'll have to integrate one curve and subtract the integral of the two curves. Just like you would if you were finding the area of the circle with integration.

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V = π∫(t² - x² + 2(r+t)√(t²-x²) + (r+t)²)dx - π∫(t² - x² - 2(r+t√(t²-x²)) + (r+t)²)dx, interval [-t, t]

Since both integrals have the same interval, we can combine them giving

V = π∫(4(r+t)√(t²-x²))dx

V = 4π(r+t)∫√(t²-x²)dx

∫√(t²-x²)dx = x√(t²-x²) + ∫x²/√(t²-x²)dx

Using methods extremely similar to my earlier "loooong solution" we can integrate the second integrand to get:

∫√(t²-x²)dx = 1/2*(x√(t²-x² + t²arcsin(x/t)) + c

integrate this from -t to t, gives -1/2*πt²

V = 4π(r+t)* -1/2*πt²

V = -2π²t²(r+t)

Volume can't be negative so you ignore it, V = 2π²t²(r+t). This is the answer I gave using the cylinder method.

longer.png

Edited by Desy ♫
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OMG... that is very complicated :P I hate integration now, but we are going to take series&differential equations option :)

btw,

∫√(t²-x²)dx = x √(t²-x²) -2 ∫x²/ √(t²-x²)dx

Integration by parts,

u=√(t²-x²)

du= -2x/√(t²-x²)

dv=dx

v=x

therefore ∫√(t²-x²)dx = x√(t²-x²) - 2 ∫x²/ √(t²-x²)dx

Thank you btw!

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Don't worry, you'll never find integration like that in the exams...

u = (t²-x²)^(1/2)

du/dx = 1/2*(t²-x²)^(-1/2)*(-2x) = -x/√(t²-x²) The twoes cancel.

∫1*√(t²-x²)dx = x*√(t²-x²) - ∫x * -x/ √(t²-x²)dx = x√(t²-x²) + ∫x²/√(t²-x²)dx

You're welcome. This was really fun to do. There's probably an easier & faster way to integrate √(t²-x²). I'll post it if I find it. This long method was more tempting as I'd already done something very similar earlier.

Edited by Gene-Peer
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oh my God.. I keep missing out the ½ and the minus sign in ∫udv = uv - ∫vdu XD haha thank you for explaining it again!!

Btw why is integrating x²/√(t²-x²) very complicated (like you need to separate it into 3 integrals first, modify it and finally integrate)?

Is there any simpler method? I prefer integrating by substitution or by part twice rather than doing that :S but if it's even more complicated then never mind :P

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Found a less complicated method. Using the awkward substitution I learnt from Wolfram|Alpha, x=t*sin(u), the integration of ∫√(t²-x²)dx is done in about 9 steps. No integration by parts needed :S Other than that, I can't think of an easier way.

Integrating x²/√(t²-x²) is generally complicated. Using the simple substitution u=x^2, gives a complicated overall method. Using the complicated (or should I say unconventional) substitution x=t*sin(u), gives a simple overall method.

Edited by Gene-Peer
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what about this:

from ∫√(t²-x²)dx = x√(t²-x²) + ∫x²/√(t²-x²)dx

∫x²/√(t²-x²) dx

Integration by substitution:

u=√(t²-x²)

du= -x/√(t²-x²)

u²=t²-x²

x=√(t²-u²)

∫x²/√(t²-x²) dx = ∫-√(t²-u²) du

So I can say that it's ∫-√(t²-x²) dx right?

Therefore:

∫√(t²-x²)dx = x√(t²-x²) - ∫√(t²-x²)dx

2∫√(t²-x²)dx=x√(t²-x²)

∫√(t²-x²)dx=x√(t²-x²)/2

But my V would be 0 :P

V=4π(r+t) * (x√(t²-x²)/2)|t-t

I get √0 there which makes everything becomes 0 :S

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heh? yes that's what I did :)

I thought that method would have worked because we just did a similar integration like that :P (getting the first integral and moving it to the other side)

but never mind, I finally used substitution of x=t sinu and found the volume :) yay! thank you Gene-Peer!! :S this must have wasted a lot of time of yours XD

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