integration Q Watch

-sonal-
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#1
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#1
find the area under the curve Y = 1 / 1 + (X)^0.5 between x = 0 and x = 1 using the substitution u = 1 + (x)^0.5 :confused:
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nota bene
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#2
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Well, what is the problem? what do you get for du/dx? what does that transform the original integral to?
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bruceleej
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#3
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What have you done so far?
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-sonal-
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#4
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(Original post by nota bene)
Well, what is the problem? what do you get for du/dx? what does that transform the original integral to?
im fine with the substitution but then i get 1 / (2u^2 - 2u) du which i dont know how to integrate??
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bruceleej
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#5
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split it up into two fractions
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nota bene
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#6
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(Original post by -sonal-)
im fine with the substitution but then i get 1 / (2u^2 - 2u) du which i dont know how to integrate??
I don't get that after the substitution.
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sam1990sam
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#7
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(Original post by -sonal-)
im fine with the substitution but then i get 1 / (2u^2 - 2u) du which i dont know how to integrate??
when substing you should get
INt 2-2/u du
Spoiler:
Show
=2u-2ln u +c
=2X^1/2 -2ln(x^1/2 + 1) +c+(2)
applying the limits
you will get 2-2ln2
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-sonal-
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#8
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how did u get that from the substitution i think i may have done the wrong substitution
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sam1990sam
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#9
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u=1+x^1/2
x^1/2=u-1
x=(u-1)^2
x=u^2 + 1 -2u
dx/du= 2u-2
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sam1990sam
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#10
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#10
int 1/(1+x^1/2) dx
= int (1/u)(2u-2) du
=int (2u-2)/u
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-sonal-
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#11
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thanx!!! i finally get it now
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sam1990sam
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#12
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(Original post by -sonal-)
thanx!!! i finally get it now
welcome
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