Help needed on complicated differentiation and integration

[wh666-666]

Hi, I just wanted to ask someone for urgent help as I am doing a computing degree which has a mathematics part to it. The question im stuck with is in regards to complex differentiation and integration. Ive tried asking mathematic teachers locally etc etc but no-one seems to be able to help so i got reccomended here so this is my first post. Below is a link to the pdf with the questions. Im fine on question 5 but im having trouble with question 4 and 6. Thanks in advance for any help in explaining it or helping with workings so i understand it.

 

http://www.geocities.com/joebloggs7654/this.pdf

[woelen]

Ive tried asking mathematic teachers locally etc etc but no-one seems to be able to help so i got reccomended here so this is my first post.

Of course they are able to help you, but these are really basic things. You should be able to find out yourself. But... you're lucky. I had a good day and will try to give you some hints which may help you .

 

Just a simple hint: All questions of (4) boil down to integration of all separate terms and adding these indefinite integrals together. You just need to know the primitive functions for all of these.

 

I'll give you one of these.

 

g(x) has two terms, one being 19/x, the other being 15x². you need to know the primitives of 1/x, and of x². These are (for x > 0), ln(x) and ⅓x³ (and having a non-determined constant term). Multiply with the factors 19 and 15, and you find your answer: 19ln(x) + 5x³ + C, where C is a indefinite constant.

 

For integration of h(u) you need to write sin²(u/10) in terms of cos(u/5), and then you integrate the function. You only are left with a cosine without higher powers.

 

Question © seems really trivial to me, I do not see why you can't solve that.

 

Question 6a: sin(3x)dx/(2+cos(3x)) = -⅓*d(cos(3x))/((2+cos(3x)). Use a change of variable z = cos(3x) and your integral becomes simple again. You only need to integrate 1/(2+z).

 

Questoin 6b: Use a similar trick as in 6a.

 

(exp(-x) - exp(x))dx = -d(exp(x) + exp(-x))

 

Now use a change of variable: z = exp(x) + exp(-x). Your differential equation now becomes:

 

dy = (-4√y * dz)/z²

 

This should not be difficult to solve anymore.

[wh666-666]

Many thanks for that as it is extremely helpful and useful for me. Is it possible for someone to explain the workings for 4b. as I am still stuggling to understand and could just do with a push in the right direction on this one.

[the tree]

4B.

[math]\int_{2}^{4}6x(2x^2 -1)\cdot dx[/math]

Well that's not to easy to integrate so expand it out.

[math]\int_{2}^{4}12x^3 - 6x\cdot dx[/math]

Then integrate it, term by term.

[math][( \int 12x^3 \cdot dx ) - ( \int 6x \cdot dx )]_{2}^{4}[/math]

[math][(12\times \frac{1}{4}x^4) - (\int 6x \cdot dx)]_{2}^{4}[/math]

[math][ 3x^4 - (\int 6x \cdot dx)]_{2}^{4}[/math]

(et cetera)

 

Once you've got that, call what you have in the square brackets [math]F(x)[/math] or whatever and work out [math]F(4)-F(2)[/math] (you may need a calculator handy), that's your awnser.

 

Oh, and just in case you really need it.[hide]The awnser is: 684[/hide]

[Tom Mattson]

joebloggs7654

 

A former Princeton Review student I take it?

[wh666-666]

Thank-oo woelen and big thankyou to tree (as tree recommended me from webdeveloper to the kind peeps on here). Thats completely solved it and i understand it all now. Seriously woelen i asked ex mathematics teachers and they looked at me blankley **eyes glazed over** thats why i came on here (education standards in UK not that high and mathematic teachers not that knowledgable). Anyway thankyou tree for the final part (it takes me a while with mathematics, not my strongest point) but i understand all the workings so its saved me alot of shred hairs!!

 

LOL by the way tom, i had to make up a geocities username on the spot and joebloggs popped in to my head (like john smith).

[Tom Mattson]

I used to teach for TPR, and I've been thoroughly indoctrinated in the "Joe Bloggs Method".

[caseclosed]

(education standards in UK not that high and mathematic teachers not that knowledgable)

so what grade are you in since you are doing this?

[the tree]

He's already said that he's at degree level, presumably he just hasn't studied maths at an elective level before.

[wh666-666]

Yea quite right tree. Done a/s-level mathematics but that didnt have any of this kinda stuff in. I decided to do a degree in computing (but it has several components to it including related mathematics) while working so they could pay for it but because i work im doing it through distance learning so there isnt as much support as it would be attending university full time.

[caseclosed]

Yea quite right tree. Done a/s-level mathematics but that didnt have any of this kinda stuff in. I decided to do a degree in computing (but it has several components to it including related mathematics) while working so they could pay for it but because i work im doing it through distance learning so there isnt as much support as it would be attending university full time.

oh, because it looks like the Calculus 2 stuff I did this year, thats why I was wondering.

[wh666-666]

Yea no probs caseclosed. I was just glad people in this thread were nice and helped me out because some of the posters in Other threads that ive been in on this site seem as if they've been eating too much infected beef. So once again many thanks for all help with workings.