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Assignment No. 2

   MTH 101 (Spring 2013)


 Maximum Marks: 15

Due Date:  9th July, 2013  

DON’T MISS THESE Important instructions:

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Lectures 30 to 35.

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Question#1                                                                                                            Marks 7


Evaluate the integral

Question#2                                                                                                            Marks 8

 (Note: In order to get full marks, do all necessary steps)


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Replies to This Discussion

Please Discuss here about this assignment.Thanks

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

Q2 ans

pt of intrsctn of graph is at +ve x-axis is at 1 & intrvls are i.e., value of x lies b/w [0,1]

use volume formula in intgral wth gvn intrl n hve ans

Best of luck

I hope now it wil b easy for all of You

Draw Graph othrwise instrctr kiya pta k kahien ka gusa kahein na utar dein aur no ap k kaat dy

Rab jany k Volume of cylndr parhaya b hy ap ko ya drct questn dy diya

go through lec 33+ n chk for volumes of cylndr

Remembr ALL Muslims in ur pryrs along with non muslims that ALMIGHTY ALLAH put them all on right path

given intrvls are on x axis [0,1], now how can we make a graph without having y axis values bcz 0,1 lies only on x axis makes no sense of any graph without y axis values

ara y baba y=0 ye intrvls intgral ki upr lower limits hain n ths is cylindricl shell case so woh frula use karein just ye crcle ka nai

bst of luck

The Volume Of Solid ...Obtained By Rotating About y-axis the region under curve y=f(x) from a->b, is given above

As it Soundz tht "The point Of intersection at +ve x-axis is at x=1(bcz as x->1 thn y->0) so it will intersect at x=1) So interval is [0,1] mean x varry from 0 to 1 or in other words b=1 & a=0"

I hope this will help little bid

Best Of Luck

note that is in the interval of integration and so, if we break up the integral at this point we get,

After getting rid of the absolute value bars in each integral we can do each integral.  So, doing the integration gives,

This assignment is easy 

g v.easy

Idea for

Q 2 i thnk to eq. r

y=1-x^2 & y=0(bcz at x-axis y=0)

so on solvng 0=1-x^2  =>  x^2=1    =>    so x = 1 & x=-1  => y=0 now furthr solve by ur self

best of luck

kya ye 100 % correct hai kya ???????????

Y=0 Now y ki dosary value kiya hogi?


y = 1-x^2 = 1-1 = 0

now @x = -1

y = 1-x^2 = 1-(-1)^2 = 1-1 = 0

y is again equal to 0

how can it possible???


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