Mastering Definite Integrals for Solving Tricky Problems

[sinClair]

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...

 

[rocomath]

Complete the square in both numerator and denominator, then use a u-sub for the radican.

 

[HallsofIvy]

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]

 

[sinClair]

Thank you so much. I was getting confused with changing the limits of integration back and forth but got it now, thanks.