Aus Online Mathematik Brückenkurs 2

The double inequality means that we look for the area of the region which is bounded above in the y-direction by the straight line \displaystyle y=x+\text{2} and from below by the parabola \displaystyle y=x^{2}.

If we sketch the line and the parabola, the region is given by the region shaded in the figure below.

As soon as we have determined the \displaystyle x -coordinates of the points of intersection, \displaystyle x=a and \displaystyle x=b, between the line and the parabola, we can calculate the area as the integral of the difference between the curves' \displaystyle y -values:

Area= \displaystyle \int\limits_{a}^{b}{\left( x+2-x^{2} \right)}\,dx

The curves' points of intersection are those points which lie on both curves, i.e. which satisfy both curves' equations:

\displaystyle \left\{ \begin{matrix} y=x+\text{2} \\ y=x^{2} \\ \end{matrix} \right.

By eliminating \displaystyle y, we obtain an equation for \displaystyle x,

\displaystyle x^{2}=x+2

If we move all x-terms to the left-hand side,

\displaystyle x^{2}-x=2

and complete the square, we obtain

\displaystyle \begin{align} & \left( x-\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}=2 \\ & \left( x-\frac{1}{2} \right)^{2}=\frac{9}{4} \\ \end{align}

Taking the root then gives that \displaystyle x=\frac{1}{2}\pm \frac{3}{2}. In other words, \displaystyle x=-\text{1} and \displaystyle x=\text{2}.

The area of the region is now given by

\displaystyle \begin{align} & \text{Area}=\int\limits_{-1}^{2}{\left( x+2-x^{2} \right)}\,dx=\left[ \frac{x^{2}}{2}+2x-\frac{x^{3}}{3} \right]_{-1}^{2} \\ & =\frac{2^{2}}{2}+2\centerdot 2-\frac{2^{3}}{3}-\left( \frac{\left( -1 \right)^{2}}{2}+2\centerdot \left( -1 \right)-\frac{\left( -1 \right)^{3}}{3} \right) \\ & =2+4-\frac{8}{3}-\frac{1}{2}+2-\frac{1}{3} \\ & =\frac{9}{2} \\ \end{align}