Area of a region bounded by $y=8$ and $y= |x^2-1|$

Question

What will be the area of the region bounded by $y=8$ and $y=|x^2-1|$?

I am having problem in plotting the graph for area required and the limits for the integral.

Answer 1

Hint:

You can divide the region into three parts:

$R_1$ where $x\le-1$ and $x^2-1\ge 0$ and $|x^2-1|=x^2-1$.

$R_2$ where $-1< x< 1$ and $x^2-1<0$, so $|x^2-1|=-(x^2-1)=1-x^2$.

$R_3$ where $x\ge 1$ and $x^2-1\ge 0$, so $|x^2-1|=x^2-1$

The points where both graphs meets are those where $8=|x^2-1|$, i.e. $$x^2-1=8\qquad \text{or}\qquad x^2-1=-8$$

New

It is easy to find the roots: $x=\pm3$.

\begin{align*} \text{Area}&=\int_{-3}^{-1}\left[8-(x^2-1)\right]dx+\int_{-1}^1\left[8-(1-x^2)\right]dx+\int_1^3\left[8-(x^2-1)\right]dx\\ &=\int_{-3}^{-1}\left(9-x^2\right)dx+\int_{-1}^1\left(x^2+7\right)dx+\int_1^3\left(9-x^2\right)dx\\ &=2\int_0^1\left(x^2+7\right)dx+2\int_1^3\left(9-x^2\right)dx\qquad\text{by symmetry}\\ &=2\left.\left[\frac{x^3}3+7x\right]\right|_0^1+2\left.\left[9x-\frac{x^3}3\right]\right|_1^3\\ &=2\left(\frac13+7-0\right)+2\left[27-9-\left(9-\frac13\right)\right]\\ &=\frac{100}3 \end{align*}

Answer 2

Can you plot $y = x^2 - 1$?

The abolute value will reflect the portion of the plot that would be below the x axis to above the x axis.

Here is a picture.