# integration QWatch

#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
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10 years ago
#2
Well, what is the problem? what do you get for du/dx? what does that transform the original integral to?
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10 years ago
#3
What have you done so far?
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#4
(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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10 years ago
#5
split it up into two fractions
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10 years ago
#6
(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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10 years ago
#7
(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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#8
how did u get that from the substitution i think i may have done the wrong substitution
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10 years ago
#9
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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10 years ago
#10
int 1/(1+x^1/2) dx
= int (1/u)(2u-2) du
=int (2u-2)/u
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#11
thanx!!! i finally get it now
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10 years ago
#12
(Original post by -sonal-)
thanx!!! i finally get it now
welcome
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