91Ó°ÊÓ

Rewrite the given equations as follows: 1. \(|x+y|=|y|-x\). We can split this into two cases: Case 1: If \(x+y \ge 0\), the equation becomes \(x+y=y-x \Rightarrow x=0\). Case 2: If \(x+y < 0\), the equation becomes \(-(x+y)=y-x \Rightarrow 2x-y=-y\). 2. \(y \geq x^{2}-1 \Rightarrow y=x^2-1\).

Plot the \(x=0\) line and \(y=2x\) line, and also the parabola \(y=x^2-1\). Observe their intersection points and the area bounded by them.

A. Intersection of \(x=0\) and \(y=x^2-1\): Set \(x=0\), we get \(y=(-1)^2-1=0\). So there is an intersection at \((0,0)\). B. Intersection of \(y=2x\) and \(y=x^2-1\): Set \(x^2-1=2x\), we get \(x^2-2x-1=0\). Calculate the roots of the quadratic equation using the quadratic formula: \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). For our case, \(a=1\), \(b=-2\) and \(c=-1\). So, the roots are \(x=\frac{2\pm\sqrt{(-2)^2-4(1)(-1)}}{2(1)}=\frac{2\pm\sqrt{8}}{2}\). So, there are intersection points at \(\left(\frac{2+\sqrt{8}}{2}, 1+\sqrt{2}\right)\) and \(\left(\frac{2-\sqrt{8}}{2}, 1-\sqrt{2}\right)\).

We can calculate the area using the integral of the difference between the upper curve and the lower curve over the interval defined by the points of intersection. The area is given by: \(A=\int_{\frac{2-\sqrt{8}}{2}}^{\frac{2+\sqrt{8}}{2}} [(2x)-(x^2-1)]dx\) First, simplify the integrand: \(2x-x^2+1\) Now, integrate with respect to \(x\): \( \int(2x-x^2+1)dx=[x^2-\frac{1}{3}x^3+x]_{\frac{2-\sqrt{8}}{2}}^{\frac{2+\sqrt{8}}{2}}\) Then, evaluate the integral at the limits of the integration and subtract: \(A=\left[\frac{1}{3}\left(\frac{2+\sqrt{8}}{2}\right)^3+\left(\frac{2+\sqrt{8}}{2}\right)^2+\left(\frac{2+\sqrt{8}}{2}\right)\right]-\left[\frac{1}{3}\left(\frac{2-\sqrt{8}}{2}\right)^3+\left(\frac{2-\sqrt{8}}{2}\right)^2+\left(\frac{2-\sqrt{8}}{2}\right)\right]\) After careful calculation, the area is approximately \(3.247\) square units.