Advanced Higher Maths 2011-2012 : Discussion and Help

Original post by soup

Are there also restrictions for the sine of the inverse sine of x? Those being -1 and 1?

No that explicit restriction is unnecessary, since it is impossible to obtain the inverse sine of anything outside of that range anyway (and still get a real number result). On the other hand, the restriction on inverse sine of sine is necessary since it's perfectly fine to obtain sines of values outside of -pi/2 to pi/2.

(edited 12 years ago)

Reply 202

Could someone do me a favour and integrate this for me quickly, just so I can make sure I'm doing it correctly?

(4x) / (2x-1)^1/2 dx

I'm getting 4(2x-1)^1/2 + C but the answers are saying different and I'm confused! :frown:

Thanks

Reply 203

Beth1234

Original post by Quintro

Thanks

Do the answers say

\frac{4}{3}(2x-1)^{1/2}(x+1)+C

by any chance, or something like that?

(edited 12 years ago)

Reply 204

Original post by Quintro

Thanks

First off, take the 4 outside. Always make the coefficient of the numerator 1.

So this becomes:

4*\int \frac{x}{(2x-1)^{1/2}}

Then let u = 2x-1 and du = 2dx

Let me know if that helps. (I realise now that it was silly to factor out the 4 to begin with; try factoring out only 2 to begin with so your numerator is 2x.)

Original post by Beth1234

Do the answers say

\frac{4}{3}(2x-1)^{1/2}(x+1)+C

by any chance, or something like that?

That's the answer I worked to. So I hope they do... :biggrin:

(edited 12 years ago)

Reply 205

Beth1234

Original post by JordanR

That's the answer I worked to. So I hope they do... :biggrin:

That's reassuring :wink:

I substituted

u=(2x-1)^{1/2}

though. Still trying to work it out the way you did it :tongue:

EDIT: Ah. Got it. Switching from scribbling on Paint to working it out with a pencil really does make a difference...

(edited 12 years ago)

Reply 206

soup

For that question would you have to simplify it down by taking out (2x-1)^1/2 as a common factor or could you just leave it as:

2/3(2x-1)^3/2 + 2(2x-1)^1/2

Reply 207

Beth1234

Original post by soup

For that question would you have to simplify it down by taking out (2x-1)^1/2 as a common factor or could you just leave it as:

2/3(2x-1)^3/2 + 2(2x-1)^1/2

I think you should take out the common factor. It looks a lot neater afterwards!

...and don't forget the constant! :tongue:

Reply 208

Original post by JordanR

First off, take the 4 outside. Always make the coefficient of the numerator 1.

So this becomes:

4*\int \frac{x}{(2x-1)^{1/2}}

Then let u = 2x-1 and du = 2dx

Let me know if that helps. (I realise now that it was silly to factor out the 4 to begin with; try factoring out only 2 to begin with so your numerator is 2x.)

That's the answer I worked to. So I hope they do... :biggrin:

Ahh! It was a completely daft mistake on my part! Sorry for wasting your time, but the help was great anyway if that's any consolation! Apparently the derivative of 2x-1 is 2x... *facepalm*. I looked over my working like four times and missed it every time. :frown:

I have a question about factoring out though. Can any number be factored out before integrating, as long as I don't take out a variable such as x?

God I'm an idiot.. :redface:

EDIT:

Original post by Beth1234

And yes, that's the answer, thanks. :smile:

(edited 12 years ago)

Reply 209

Original post by Quintro

I have a question about factoring out though. Can any number be factored out before integrating, as long as I don't take out a variable such as x?

God I'm an idiot.. :redface:

EDIT:

And yes, that's the answer, thanks. :smile:

It happens to everyone! Don't worry about it. I remember being stuck with a question and then realised that I was raising the power by one and then multiplying by it, in some sort of pseudo blend of integration and differentiation, and wondered why I was getting a crazy answer.

Yes. Any constant can be factored out. That can make trig integrals a lot, lot simpler.

Original post by Beth1234

That's reassuring :wink:

I substituted

u=(2x-1)^{1/2}

though. Still trying to work it out the way you did it :tongue:

EDIT: Ah. Got it. Switching from scribbling on Paint to working it out with a pencil really does make a difference...

Ah, good. Saves me Latexing my answer...

Reply 210

I did make a stupid mistake earlier but I'm still not getting this.. I'd appreciate if someone could look through my working to see where I'm screwing up...

EDIT: I know I made an error at the bottom, should be a minus sign not a plus.

Reply 211

Original post by Quintro

I did make a stupid mistake earlier but I'm still not getting this.. I'd appreciate if someone could look through my working to see where I'm screwing up...

EDIT: I know I made an error at the bottom, should be a minus sign not a plus.

Mistake is going from third-to-bottom line to second-to-bottom. Taking out 4/3 from (4x - 2)/3 should take out a 4 from the 2 as well, which leaves 1/2.

(edited 12 years ago)

Reply 212

Original post by Quintro

I did make a stupid mistake earlier but I'm still not getting this.. I'd appreciate if someone could look through my working to see where I'm screwing up...

EDIT: I know I made an error at the bottom, should be a minus sign not a plus.

Third to second last line is incorrect.

(4x-2)/3 + 2 can be rewritten as

\frac{4(x+1)}{3}

That should help. :smile:

Edit: damn it. Beaten. :tongue:

Reply 213

Original post by ukdragon37

...

Original post by JordanR

...

Thanks very much guys, I've got it now. I really need to work on manipulating equations. :tongue:

And I have no idea why I'm doing Maths at this time :s-smilie:

Reply 214

Original post by Quintro

Thanks very much guys, I've got it now. I really need to work on manipulating equations. :tongue:

And I have no idea why I'm doing Maths at this time :s-smilie:

It's what learning calculus did for me most. Remember that in an exam, they wouldn't penalise you for not putting it into that form (unless they asked you to show that it equalled that specifically).

So don't worry about it. c:

Maths at this time? Well, it's because maths is great. Or because you need something to bore you to sleep.

Reply 215

Original post by nerd434

I´m only in fourth year just now studying Intermediate 2 Maths, currently I can just repeat questions out a textbook to understand a concept and move onto the next. Is this the same with Higher and Adv Higher? Is it all about practice?

And what is everyone aiming for in their exam?

It's similar for Higher. However for AH you are more likely to get some question whose style you have never seen before and you need to work out what method to use on it. Of course practising as much as possible helps. You also need to have a good memory as there are many formulae and identities to remember.

Reply 216

Original post by ukdragon37

What I've noticed is that there's often more than one method that you can use, too. You demonstrated that the other day with some differentiation.

Reply 217