Let us first find the bounds of integration i.e. the points where the two points intersect each other. To do that we need to solve the following system of equations.

`y=x^2-6`

`y=0`

Substituting the second equation into the first yields

`x^2-6=0`

`x^2=6`

`x_(1,2)=pm sqrt6`

If we now look at the...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Let us first find the bounds of integration i.e. the points where the two points intersect each other. To do that we need to solve the following system of equations.

`y=x^2-6`

`y=0`

Substituting the second equation into the first yields

`x^2-6=0`

`x^2=6`

`x_(1,2)=pm sqrt6`

If we now look at the image below, we see that the whole region is below `x`-axis. This means that the integral will be negative so to calculate the area we need to put the minus sign in front of the integral (we could also put the integral in absolute value or we could simply switch lower and upper bounds with each other).

`A=-int_-sqrt6^sqrt6 (x^2-6)dx`

Let us now calculate the area of the region.

`A=-(x^3/3-6x)|_-sqrt6^sqrt6=-((6sqrt6)/3-6sqrt6+(6sqrt6)/3-6sqrt6)=8sqrt6`

**Therefore, the area of the region bounded by the parabola `y_1=x^2-6` and the line `y_2=0` is `8sqrt6.` **