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Find the area of the region bounded by the curves $y=x^2$ and $y=x$.

Find the area of the region bounded by the curves $y=x^2+1$ and $y=2$

I have a ton of questions like this and I have been graphing them and then splitting them into intervals and adding them up but this is giving me an answer thats a little off and its taking there a faster way? Also I am stuck on $y=x^2+1$ and $y=2$ because I dont know what region they want..I see $y=2$ as a line intersecting $x^2+1$, when I graph it.

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HINT They ask for the area of the yellow region: enter image description here

The areas would be given by integrals $\int_{x_1}^{x_2} \left(y_\text{top}(x) - y_\text{bottom}(x)\right) \mathrm{d} x$ with appropriate choices of boundaries $x_1$ and $x_2$ and functions $y_\text{top}(x)$ and $y_\text{bottom}(x)$.

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your curves are nice, what software do you use? – NoChance Nov 30 '11 at 21:12
@EmmadKareem I am using Mathematica (site). – Sasha Nov 30 '11 at 21:13
pretty nice, thanks. – NoChance Nov 30 '11 at 21:16

The fastest way to find the area is to use integration. The area is the result of definite integral of the difference between the two functions.

The area bounded by $y=x^2+1$ and $y=2$ is shown below:

enter image description here

It is the area between the curve and the line of course.

First, we have 2 points that we need to find. The points are those of the intersection of the line and the curve. You can find the points by solving the equation:

$y=x^2+1$ and $y=2$.

You get

$y=x^2+1=2$ and this gives you the points:

$(-1,2)$ and $(1,2)$.

The area A is:

$A=\int_{x=-1}^{x=1} 2-(x^2+1) dx =4/3$.

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