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Calculate the area of the region of the graph bounded by:

$$\begin{eqnarray} y &=& x \\ y &=& x^2 + 1 \\ y &=& 2 \\ x &=& 0 \end{eqnarray}$$

My final result is $\displaystyle\frac 83.$

For the first function I trivially found $x = 2.$

For the second, I did the equation $x^2 + 1 = 2,$ which gave me $x = -1$ and $x = 1,$ but since the inferior limit is $0,$ $x = 1$ is the only solution.

So I calculated the integral of $(x^2 + 1 - x)$ from $0$ to $1,$ and the integral of $(x^2 + 1 - x)$ from $1$ to $2,$ and the result was $\displaystyle\frac 83,$ but I'm not so sure I did everything right. Can you please tell me what I did wrong in this problem?

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  • $\begingroup$ You seem to be assuming these four equations are to be satisfied with one value of $(x,y)$, which is not possible. Each of the for represents a curve in the plane. You can solve (some) pairs of the as simultaneous equations to find the crossing points of he curves. $\endgroup$ – Ross Millikan Oct 30 '14 at 20:16
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If you plot all functions involved, then the area is given by

$$\int_0^1(x^2+1-x)dx+\int_1^2(2-x)dx.$$

region

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