Math Help - Area from a definite integral

1. Area from a definite integral

I am suppose to obtain an area bound by 3x + y = 6 above and below by y = x^2 - 4 and the y axis. I think I should take the integral of the first and subtract the integral of the second one from 0 to 2

In terms of x: 6x - 3x^2/2 - (x^3/3 + 4x) from x = 0 to 2. Am I on the right track on this one????? Please let me know. I asked someone else in my class and they added the two integrals and used |x^2 - 4| for second integral?? Thanks again for this wonderful site!!!! Frost King

2. to find the area between the line and the curve, first find where the line and the curve intersect.

$y = 6 - 3x$

$y = x^2 - 4$

$x^2 - 4 = 6 - 3x
$

$x^2 + 3x - 10 = 0$

$(x + 5)(x - 2) = 0$

$x = -5$ , $x = 2$

$A = \int_{-5}^2 (6-3x) - (x^2 - 4) \, dx$

simplify the integrand, integrate, and use the FTC.