Need Someone to check my Differentiation & Integration Asgmt

[NiCeBoY]

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

The maximum limit for one post is set to 3 attachment.. i will make another post for more 3 pages for my answers..





Thanking you in advance...

 

[NiCeBoY]

Here are my answer pages 3 - 5..

thx

 

[Deleted member 4993]

for question 4(a)

Graph it in your calculator.

You'll find there are 3 points of intersections (0, 1 , 2).

Also look carefully which graph produces upper-bound (hence larger area and that will avoid getting negative answer for area) and in which domain.

for 3

use

u = 1 + x^4

 

[NiCeBoY]

Hello,
my calculator does not support graph bro.. how do i proceed then?

Also have you checked the other questions.. are they good..?

thx man..

 

[NiCeBoY]

For Question 4(a) and question 3(first part) & question 1(3) its ok i got the answer..thx..

Can anyone how to answer question 2(b) and 4(b) please?

Thanks..

 

[mmm4444bot]

Can anyone show how to answer question 2(b) ...

Are you sure that you're asking about 2(b)? I already see an answer for 2(b) on your paper.

By the way, x < 1.5 may be close enough for your instructor, but I would report it differently.

0 < x < 1.5

~ Mark

 

[NiCeBoY]

thanks man that what i wanted to know if x<1.5 is enough or beter to put it as 0<x<1.5..

Anyway what about question 4(b) can you help me how to proceed please?

thanks.

 

[mmm4444bot]

thanks man that what i wanted to know if x<1.5 is enough or beter to put it as 0<x<1.5.

You are most welcome. I always enjoy learning that I inadvertently answered a question ...

... what about question 4(b) can you help me how to proceed please?

I cannot help you anytime soon; I hope somebody else does, shortly. In the meantime, it would be a big help if you could explain what YOU know about this problem. (I'm just returning from a four-year absense from mathematics, and the method for this problem is very fuzzy in my mind ... heh, heh ... that's the worst kind of "fuzzy math"!)

If I were motivated to refresh my memory, then I would start by reading about revolution of solids.

Click here for Google results on search string "volume solid revolution calculus"

~ Mark

 

[NiCeBoY]

Thanks man.. i got the answer..
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..

 

[Deleted member 4993]

\(\displaystyle \int (2x+1) \, cos(x) \, dx\)

You know

\(\displaystyle \int u dv \, = \, u\cdot v \, - \, \int v du\)

here

u = 2x + 1
du = 2 dx

and

dv = cos(x) dx
v = sin(x)

then

\(\displaystyle \int (2x+1) \, cos(x) \, dx\)

\(\displaystyle = \, (2x+1)\cdot sin(x) \, - \, \int sin(x) \, [2 \, dx]\)

Now continue....

 

[NiCeBoY]

up to here I am ok...I can work..
but how do i solve :

What i know is somewhere i should get: du=dx

but i cant get it out..

 

[Deleted member 4993]

up to here I am ok...I can work..
but how do i solve :

What i know is somewhere i should get: du=dx

but i cant get it out..

\(\displaystyle \int sin(x)[2 \, dx]\)

\(\displaystyle = \, 2\cdot\int sin(x) \, dx\)

Now does it look familiar....

 

[NiCeBoY]

ohhhhhh yes...
thanks a lot
cheers!!..
thanks great..
thx again..