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NiCeBoY

Member

Registered: 2008-08-30

Posts: 6

Need Someone to check my Differentiation & Integration Asgmt

Hello,
I am new here.I got an assignement due in less then 14 hours that is tomorrow morning..
I am posting the questions and my answers and am lacking behind 3 numbers.

Can anyone check the questions and my answers and let me know if there is anything wrong please? and also help me out quick with the 3 numbers that am not able to get a solution.

I have attached to this post the questions and my answers.

Questions.JPG = Contain the questions.
    Page10001.JPG - Page10005.JPG = Contain my answers

Thanking you in advance...

Last edited by NiCeBoY (2008-08-30 14:12:28)

NiCeBoY

Member

Registered: 2008-08-30

Posts: 6

Re: Need Someone to check my Differentiation & Integration Asgmt

here are the ques and ans..

- The question paper.

- Page 1 My Answers

- Page 2 My Answers

- Page 3 My Answers

- Page 4 My Answers

- Page 5 My Answers

thx

Last edited by NiCeBoY (2008-08-30 14:17:47)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Need Someone to check my Differentiation & Integration Asgmt

For Q1(3), use the chain rule.

For Q3(1), let

.

In Q4(1), the cubic curve lies below the straight-line curve between your limits of integration. hence you must subtract the cubic curve from the stright-line curve, not the other way round.

NiCeBoY

Member

Registered: 2008-08-30

Posts: 6

Re: Need Someone to check my Differentiation & Integration Asgmt

for Q3(1)

then how do i proceeed?

then how i substitute it please?

Last edited by NiCeBoY (2008-08-30 17:25:12)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Need Someone to check my Differentiation & Integration Asgmt

---

Why did the vector cross the road?
It wanted to be normal.

NiCeBoY

Member

Registered: 2008-08-30

Posts: 6

Re: Need Someone to check my Differentiation & Integration Asgmt

thanks mathsyperson i finally got the answer and understood the situation..

Can you please check Question1(3) for me if possible please? am struggling with...

Thx..

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Need Someone to check my Differentiation & Integration Asgmt

As Jane said, use the chain rule.

Set t = x² + log x.

Then dy/dt = 1/(2√t) and dt/dx = 2x + 1/x.

Therefore dy/dx = dy/dt * dt/dx = (2x + 1/x)/[2√(x² + log x)].

---

Why did the vector cross the road?
It wanted to be normal.

NiCeBoY

Member

Registered: 2008-08-30

Posts: 6

Re: Need Someone to check my Differentiation & Integration Asgmt

When i simplify it :

2x/[x+2√(x² + log x)]

Is my simplification right?

And what about for Question 2.

how do i answer part (b) What are the possile values of x.

i think the answer is : 0< X < 1.5

Am i rite plz?

thanks for ur help man.

NiCeBoY

Member

Registered: 2008-08-30

Posts: 6

Re: Need Someone to check my Differentiation & Integration Asgmt

Thanks man.. i got the answer for all the questions..
but am getting some issues with question 3 (second part)

Can you please do it step by step it seems i got some errors in..
thanks in advance..

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Need Someone to check my Differentiation & Integration Asgmt

Let

Let

You need to make u = something and dv = something, not u = something and v = something
In general let u be complicated and let dv be something easy to integrate (stuff usually simplifies when u diff it)

Last edited by Identity (2008-08-31 23:23:27)

NiCeBoY

Member

Registered: 2008-08-30

Posts: 6

Re: Need Someone to check my Differentiation & Integration Asgmt

Thanks man
very helpful...
thanks a lot..
This post/thread can be closed:)

Problem solved.!!