MHB Area finite region bounded by the curves

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The discussion revolves around calculating the area of a region bounded by the curves y²=1-x and y=x+1. The user initially encounters difficulties and incorrectly computes the area as -9/2. After receiving advice to sketch the region for clarity, they realize the parabolic function is greater than the linear function within the integration limits. This leads to a corrected calculation, resulting in the accurate area of 9/2. The exchange highlights the importance of visualizing functions to avoid mistakes in integration.
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 alot! 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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