Integrating the square root of a linear function.

  • Thread starter cptrsn
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  • #1
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Homework Statement


[tex]\int\sqrt{2*x-1}[/tex]


Homework Equations





The Attempt at a Solution


Homework Statement


Obviously this can be fixed with the antiderivative of a linear function with proper constants i.e. [tex]\int A\times f(bx + c) = \frac{A}{b} F(bx + c)[/tex], however my instructor provided me with a "clue" which I somehow cant seem to work out.

He claims that [tex]\int\sqrt{2*x-1}[/tex] is equivilant to [tex]\int u^2[/tex] using "the right" substitution. My idea usually is:
1) [tex]u = \sqrt{2*x - 1}[/tex]
2) [tex]x = u^2 [/tex]
3) [tex]dx = 2u du[/tex]

Which would lead to [tex]\int u dx = \int u 2u du = \int 2u^2 du[/tex].
Obviously this is NOT identical to what he suggests, so can anyone point me in the right direction? I'm quite keen to know how he see such a problem.
 

Answers and Replies

  • #2
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If u = sqrt(2x-1)
then how x = u^2?
 
  • #3
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I don't think he quite knew what he was talking about; there are some problems with that which I wouldn't even try to work with or fix.
What part of the integrand can u be that will easily let you integrate with respect to u?
 
  • #4
HallsofIvy
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The teacher not being here to defend himself, I might suggest that cptrsn did not quite remember what the teacher had suggested. Any time you have a linear term, ax+ b, an obvious substitution is just u= ax+ b so that du= adx.
 

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