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Past paper 1, 2003, Question 8 I'm seeking help on integration

xkarenx

Posted 29 January 2006 - 07:58 PM

I have the prelim next week and really need some help with this integration stuff. It seems to be the one main topic I really can't remember and it's likely I'll get a question in the prelim related to integration. I actually can't remember the basic rule of how to do it and I was hoping someone could explain it and let me know the working on this past paper question so I can practise. Incase you don' have it this is the question..

user posted image

The Wedge Effect

Posted 29 January 2006 - 08:12 PM



\begin{align*}\int^1_0 {\frac {dx}{(3x+1)^{\frac12}}}\end{align*}

First of all, you move the dx to the side, so it becomes:



\begin{align*}\int^1_0 {\frac {1}{(3x+1)^{\frac{1}{2}}}} \,dx\end{align*}

Then you change it around so that:



\begin{align*}\int^1_0 {(3x+1)^{-\frac {1}{2}}} \,dx\end{align*}

You then integrate the expression and substitue the upper and lower limit for the values of x, with the lower limit's result being subtracted from the upper limit's result and you have your answer.

Edited by Wedge37, 29 January 2006 - 08:42 PM.


xkarenx

Posted 29 January 2006 - 08:19 PM

Thanks for replying! I found some notes on the website that helped remind me of the basic rules. The only thing confusing me now is why you began with the power of a half on the top rather than the bottom..



\begin{align*}\int^1_0 {\frac {dx}{(3x+1}^{\frac{1}{2}}}\end{align*}

The Wedge Effect

Posted 29 January 2006 - 08:19 PM

QUOTE(xkarenx @ Jan 29 2006, 08:19 PM)
Thanks for replying! I found some notes on the website that helped remind me of the basic rules. The only thing confusing me now is why you began with the power of a half on the top rather than the bottom..



\begin{align*}\int^1_0 {\frac {dx}{(3x+1}^{\frac{1}{2}}}\end{align*}

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That was a coding mistake. Sorry! The above post has now been edited. Thanks for pointing that out. laugh.gif

xkarenx

Posted 29 January 2006 - 08:30 PM

Oh ok.. well actually I'm still confused on how to do the question. I don't get why the 1 on the top line can disappear and nothing changes with the bottom half. I guess i'll just have to skip by any integration in the prelim and ask my teacher once we are back at school for help on the lot. Thanks for trying to help anyway.

Dave

Posted 29 January 2006 - 08:39 PM

i think that should be to the power of negative a half if you are changing it so that it is not one over




\begin{align*}\int^1_0 {(3x+1)^{\frac {-1}{2}}} \,dx\end{align*}

xkarenx

Posted 29 January 2006 - 08:43 PM

Yeah that's what I thought it probably was but still I don't know where to go from there. Maybe if I get a question like that I can at least rearrange it to that form before moving on then there's a tiny chance I could get a mark for doing something.

The Wedge Effect

Posted 29 January 2006 - 08:44 PM

Meh, I keep making mistakes with the damn LaTeX code. Sorry, that should have been to a negative power. And it's because of the Laws of Indices that the 1/<value> disappears. For example:



\begin{align*}\frac {1}{x^2} = x^{-2}.\end{align*}

xkarenx

Posted 29 January 2006 - 09:11 PM

I just marked that paper and got myself 46% ermm.gif looks like I won't be passing this prelim at all! I need 50% for a C don't I?

xkarenx

Posted 29 January 2006 - 10:21 PM



\begin{align*}\int^1_0 {(3x+1)^{\frac {-1}{2}}} \,dx\end{align*}

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This might sound really stupid but how do you multiply out the brackets to a power of negative a half? It's non calculator.

George

Posted 30 January 2006 - 09:25 AM

For questions like that, you don't multiply out the brackets. There's a rule for dealing with them that's covered in Unit 3, so you might not have met that yet smile.gif

You can see the rule, and a couple of examples, on page 4 of our Further Calculus notes.

xkarenx

Posted 30 January 2006 - 07:54 PM

Ah I see. Thanks for the link I guess I don't have to worry about that question just now then. I spent today looking over my how to pass higher maths book and other notes and started integration from the beginning again to make myself familiar. Hopefully I'll be able to do any that comes up now or if not manage it all at least get some marks from my attempts. My only other main problem at the moment would be the compound and multiple angles section. I'm never sure what to do when faced a question such as this I came across in past papers today..

Solve the equation 3cos(2 powerx.gif )+10cos( powerx.gif )-1=0 for 0 lessorequal.gif powerx.gif lessorequal.gif pi.gif, correct to 2 decimal places.

If anyone can be bothered to explain in steps how to tackle this question I'd be very grateful.

The Wedge Effect

Posted 30 January 2006 - 08:27 PM

You expand the double angle terms in that equation, so you'd have:



\begin{align*}(3\cos^2(x)-1) + 10\cos(x)-1=0\end{align*}

You then simplify the expression to get:



\begin{align*}3\cos^2(x)+10\cos(x)-2=0\end{align*}

As you can see, this is a quadratic of the form ax^2+bx+c=0

Substitue y=\cos(x)\, into the expression:



\begin{align*}3x^2+10x-2=0\end{align*}

You can then factorise this expression and then replace the y values with the cos x values. It can then be simplified to get two values of x, in radians, of course. But be cafeful of the limits of the range of values of x you solve for. In this case, it's from 0 to 

\begin{align*}\Large \pi\end{align*}


Edit: If you want me to clarify any further, feel free to ask me and I'll edit above post to final solutions and add a few more pointers. smile.gif

Edited by Wedge37, 30 January 2006 - 08:31 PM.


xkarenx

Posted 30 January 2006 - 08:36 PM

Thanks for that it makes it a lot clearer. I'll just have to try and remember to get it into the form of the quadratic equation. I'm pretty clear on the factorising and then finding the angles or radians it's usually just rearranging them and selecting the correct term to change cos(2x) or sin(2x) to. I'll just finish the question just now and let you know if I come across any other problems. Thanks again.

xkarenx

Posted 30 January 2006 - 08:48 PM

I spoke to soon. I thought this question was easier than it is. The answer at the back says '1.23 radians only' but I'm not sure how it is getting that.

The Wedge Effect

Posted 30 January 2006 - 09:07 PM

QUOTE(xkarenx @ Jan 30 2006, 08:48 PM)
I spoke to soon. I thought this question was easier than it is. The answer at the back says '1.23 radians only' but I'm not sure how it is getting that.

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Just remember the limits, from 0 to \pi That's probs why you end up with a different answer. Also, maybe you forgot to change your calculator to radians mode when calculating the answers to this questions. Values between 0 to 3.141 are correct.

Steve

Posted 30 January 2006 - 09:23 PM

Wedge, you made a little mistake in your substitution.

I've attached my solution to this question, let me know if it helps (or doesn't!).

Attached Files



The Wedge Effect

Posted 30 January 2006 - 09:30 PM

QUOTE(Steve @ Jan 30 2006, 09:23 PM)
Wedge, you made a little mistake in your substitution.

I've attached my solution to this question, let me know if it helps (or doesn't!).

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Oh whoops. Silly me. I'm no good when trying to give solutions on pc. I prefer to do it on paper, make less mistakes that way.

xkarenx

Posted 30 January 2006 - 10:20 PM

Thanks for spotting that mistake Steve it has solved my problem. I get the correct answer now. The solution was also very helpful to look through! Wedge thanks for the effort too.

The Wedge Effect

Posted 31 January 2006 - 02:37 PM

No problems...I'll just try to do it on paper first from now on. tongue.gif

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