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Oh!! try this link:
http://www.gceguide.com/Books/ebs/e...orked_Examples__Fourth_Edition-JUGG3RNAUT.pdf
this will work!
Yes It's working..Thanks a lot.
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Oh!! try this link:
http://www.gceguide.com/Books/ebs/e...orked_Examples__Fourth_Edition-JUGG3RNAUT.pdf
this will work!
no prob!Yes It's working..Thanks a lot.
No. It wouldve been easier with a substitution though.NO..nothing as such is given in the question.
No. It wouldve been easier with a substitution though.
But The reason is that d(arctan x)/dx = 1/(1+x^2)
and the integral is the opposite of the derivative
Its not possible without substitution. Take substitution y = tanx or no. I dont understand itNO..nothing as such is given in the question.
If you applied the correct identity it might help . . . tan2(x)+1=sec2(x) Your integral equation then becomes, ∫1x2+1dx=∫dθbut how is it possible to solve that with substitution ? and which substitution will be used?
Is it choice A?Can someone please help me with this question
ii)Take 2 out.PLZ help me with part 2 !!
Just the answers?
Yeah....... THNX!!!!!!Just the answers?
i) 106.3
ii) 7.2 N
Anytimee.Yeah....... THNX!!!!!!
I did the ques but I couldn't find the answers anywhere......... Just wanted to confirm........ Thnx
Awww OkiAnytimee.
i) 106.26
ii)Take 2 out.
2 ʃ ((2-u)^2 / u) du
Solve the square.
2 ʃ (4 - 4u +u^2)/u
Make fractions.
2 ʃ 4/u – 4u/u + u^2 /u)
Cancelling out will give you.
2 ʃ 4/u – 4 + u
Now integrate.
2 [ 4ln IuI – 4u + u^2/2 ]
8ln IuI - 8u + u^2
Plug in the limits.
(8ln I2I – 8(2) + (2^2)) – (8ln I1I – 8(1) + 1^2)
(8ln I2I – 16 + 4) – (0 – 8 + 1)
(8ln I2I – 12) – ( - 7)
8ln I2I - 12 + 7
8ln I2I – 5 Ans.
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