Area finite region bounded by the curves

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SUMMARY

The discussion focuses on calculating the area of the finite region bounded by the curves defined by the equations \(y^2=1-x\) and \(y=x+1\). The user initially miscalculated the area, obtaining \(-\frac{9}{2}\), but after guidance, correctly determined the area to be \(\frac{9}{2}\). The key to solving the problem was recognizing that the parabolic function \(y^2=1-x\) is greater than the linear function \(y=x+1\) over the specified limits of integration, which are \(x_1=-2\) and \(x_2=1\). Sketching the region helped clarify the integration process.

PREREQUISITES
  • Understanding of definite integrals
  • Familiarity with curve sketching techniques
  • Knowledge of polynomial and linear functions
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the method of finding areas between curves using definite integrals
  • Learn about the properties of parabolic functions and their intersections with linear functions
  • Explore graphical methods for visualizing bounded regions in calculus
  • Practice solving similar problems involving integration of bounded areas
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques and area calculations between curves, as well as educators looking for examples to illustrate these concepts.

Petrus
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Hello MHB,
I got stuck on an old exam
determine the area of the finite region bounded by the curves $$y^2=1-x$$ and $$y=x+1$$ the integration becomes more easy if we change it to x so let's do it
$$x=1-y^2$$ and $$x=y-1$$
to calculate the limits we equal them
$$y-1=1-y^2 <=> x_1=-2 \ x_2=1$$
so we take the right function minus left so we got
$$\int_{-2}^1 y-1-(1-y^2) <=> \int_{-2}^1 y+y^2-2$$ and I get the result $$- \frac{9}{2}$$ and that is obviously wrong... What I am doing wrong?

Regards,
$$|\pi\rangle$$
 
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Over the limits of integration, the parabolic function is greater than the linear function, this is why it is a good idea to sketch the region first so that you can see more clearly what you need to do. :D
 
MarkFL said:
Over the limits of integration, the parabolic function is greater than the linear function, this is why it is a good idea to sketch the region first so that you can see more clearly what you need to do. :D
Thanks a lot! I learned a lesson this time :) I did do it in my brain and that was not cleaver! Thanks a lot for the fast responed!:) Now I get $$\frac{9}{2}$$ that is correct :)

Regards,
$$|\pi\rangle$$
 

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