Confusing Definite Integral

  • Thread starter sinClair
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  • #1
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Homework Statement


Integrate [tex]\int_{0}^{1}\sqrt{\frac{4x^2-4x+1}{x^2-x+3}}dx[/tex]


Homework Equations




The Attempt at a Solution


U sub: let [tex]u=x^2-x+3[/tex] Then [tex] du=2x-1[/tex] and then have to evaluate [tex]\int_{3}^{3}\sqrt{\frac{du^2}{u}}dx[/tex] But how with these limits of integration should this be 0? Not sure how to evaluate this....
 

Answers and Replies

  • #2
1,753
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Complete the square in both numerator and denominator, then use a u-sub for the radican.
 
  • #3
HallsofIvy
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Complete the square in both numerator and denominator, then use a u-sub for the radican.
I see no reason to do that.

Homework Statement


Integrate [tex]\int_{0}^{1}\sqrt{\frac{4x^2-4x+1}{x^2-x+3}}dx[/tex]


Homework Equations




The Attempt at a Solution


U sub: let [tex]u=x^2-x+3[/tex] Then [tex] du=2x-1[/tex] and then have to evaluate [tex]\int_{3}^{3}\sqrt{\frac{du^2}{u}}dx[/tex] But how with these limits of integration should this be 0? Not sure how to evaluate this....
[itex]\sqrt{x^2}= |x|[/itex] and 2x- 1 changes sign at x= 1/2. This integral is the same as
[tex]\int_0^1 \frac{|2x-1|}{\sqrt{x^2- x+ 3}}dx= -\int_0^{1/2}\frac{2x-1}{\sqrt{x^2-x+ 3}}dx+ \int_{1/2}^1 \frac{2x-1}{\sqrt{x^2- x+ 3}}dx[/tex]
 
  • #4
22
0
Thank you so much. I was getting confused with changing the limits of integration back and forth but got it now, thanks.
 

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