Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 12

You need to determine first the points of intersection between curves y = sqrt(12-4x) and y = x , by solving the equation, such that:
sqrt(12-4x)= x =>12-4x = x^2 => x^2 + 4x - 12 = 0
x_(1,2) = (-4+-sqrt(16 + 48))/2 => x_(1,2) = (-4+-8)/2
x_1 = 2; x_2 = -6
Hence, the endpoints of integral are x = -6 and x = 2
You need to decide what curve is greater than the other on the interval [-6,2]. You need to notice that sqrt(12-4x) >x on the interval [-6,2],hence, you may evaluate the area of the region enclosed by the given curves, such that:
int_a^b (f(x) - g(x))dx , where f(x) > g(x) for x in [a,b]
int_(-6)^2 (sqrt(12-4x) - x)dx =int_(-6)^2sqrt(12-4x) dx - int_(-6)^2 xdx
You need to solve int_(-6)^2sqrt(12-4x) dx using substitution 12 - 4x = t , such that:
12-4x = t => -4dx = dt => dx = -(dt)/4
int_(-6)^2sqrt(12-4x) dx= -(1/4)int_(t_1)^(t_2)sqrt(t) dt
int_(-6)^2sqrt(12-4x) dx= -2/12 (12-4x)^(3/2)|_(-6)^2
int_(-6)^2sqrt(12-4x) dx= -2/12 (12-4*2)^(3/2) + 2/12(12 + 24)
int_(-6)^2sqrt(12-4x) dx= -(2/12)*8 + (36*2)/12
int_(-6)^2sqrt(12-4x) dx= - 4/3 + 6
int_(-6)^2sqrt(12-4x) dx= (18-4)/3
int_(-6)^2sqrt(12-4x) dx= 14/3
Hence, evaluating the area of the region enclosed by the given curves, yields int_(-6)^2sqrt(12-4x) dx= 14/3.

The area of the region enclosed by the given curves is found between the red and orange curves, for x in [-6,2].