Find the area of the region determined by the system: \begin{align} y & \ge |x| \\ y & \le -|x+1| +4 \\ \end{align}

Question

Find the area of the region determined by the system: \begin{align} y & \ge |x| \\ y & \le -|x+1| +4 \\ \end{align}

My attempt

• Assuming $x>0$

I have the system $$\begin{cases}\begin{align} y &\ge x \\ y & \le -x+3 \\ \end{align}\end{cases}$$

• Assuming $x<0$

$$\begin{cases}\begin{align} y &\ge -x \\ y & \le x+5 \\ \end{align}\end{cases}$$

A relevant interval is $-1<x<0$ because in this interval $y \le -|x+1| +4 $ is still positive while $y \ge |x|$ is negative.

How do I combine now this information to solve the problem ?

I don't see how I can get something of the form $y \ge a$ and $y \le b$ which give clear bounds about area... in this problem it seems more complicated.

Answer 1

If you can use graphs, then draw the graphs of the functions. The graph of $y=|x|$ is a pretty basic one. You can use transformation to get the graph of the other one. The following is the resulting picture:

You can see that it forms a rectangle. Can you proceed from here?

By the way, if you use algebraic way, you have to discuss three cases: $$x\leq-1\\ -1\leq x \leq 0\\ x\geq 0$$

Answer 2

Hint:You have 4 equations of form $y=mx+c$ so find $4$ points of intersection $2$ each with opposite slopes also its a polygon with $2$ pair of lines parallel . Can you proceed further.

Answer 3

For $x > 0$, your area is the area enclosed by the graphs of $y = x$ and $y = -x + 3$ between $0$ and their intersection point. Thus, determine $x = -x +3$ which gets you $2x = 3 \Leftrightarrow x = {3\over 2}$ . So the first area is $$\int_{0}^{3 \over 2} 2x - 3 dx$$

Process for $x < 0$ analogously.