Calculus of a Single Variable, Chapter 7, 7.1, Section 7.1, Problem 1

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.