Dido525
Evaluate the intergral:



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Dido525
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dw:1355802290129:dw

Dido525
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I thought I could use substitution but I can't ...

winterfez
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is this right?
\[\int\limits_{0}^{4}\frac{ x }{ \sqrt{1+2x} }\]

Dido525
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yep :) .

eliassaab
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put u= 1 + 2 x

Mimi_x3
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\( u = 1+ 2x => du/dx = 2\)
\( 1+2x = u => 2x = u1 => x = (u1)/2 \)

Dido525
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I got to that step. Then I got stuck :P .

Mimi_x3
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Maybe:\[\int\limits\frac{x}{\sqrt{1+2x}} dx => \int\limits \frac{(u1)/2}{\sqrt{u}} *2du \int\limits\frac{u1}{2} *\frac{1}{\sqrt{u}} 2du => 2\int\limits\frac{u1}{2} *\frac{1}{\sqrt{u}} \]

eliassaab
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Here what wolfram alpha gave

Mimi_x3
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Would my step work tho?

zepdrix
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@Dido525 do you understand what mimi did? those are the correct steps to take.

Dido525
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Yeah I actually do. I just need to process it.

Dido525
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@Mimi_x3 : Wouldn't you write du/2 instead of 2*du ?

Dido525
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Because we said du/2 was dx.

zepdrix
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Oh yes :o good call.

Mimi_x3
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Yeah typo lol
sorry! that was where i went wrong :(

Dido525
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It's okay :) .

Dido525
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So so far I have:
\[\frac{ 1 }{ 4 }\int\limits_{1}^{9} \frac{ u1 }{ \sqrt{u} } du\]

Mimi_x3
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then break it up!

Dido525
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Ohh... Right :P .

Mimi_x3
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integrate it; then don't forget to sub back in \(u\)

Dido525
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Thanks! :) .

Mimi_x3
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:)