Calculate the area enclosed by the curve and line Calculate the are enclosed by ${y = 2x - 1}$ and ${y=  x^2 + 6x + 2}$
First of all I combine the equations into:
${x^2 + 4x + 3 = 0}$
${(x + 3)(x + 1), x = -3,  x = -1}$
They intersect at ${(-3 -7) (-1, -3)}$
I would say that ${y = 2x  -1}$ is the top equation so to work out the area, I would use:
${\int_{-3}^{-1} (2x -1) - (x^2 + 6x + 2) dx}$
=> ${-{x^3\over 3} - 2x^2 - 3x}$
=> ${(-{1\over 3} - 2 + 3) - ({-27 \over 3} -18 + 9)}$
=> ${{2\over 3} - (-18)}$
=> ${18{2\over 3}}$
This is not the correct answer, can anyone point out where I have gone wrong?

$(\frac13-2+3)-(9-18+9)=\frac43$ when there are two equation, there are two roots.  You should use
$$\int_{α}^{β}a(x-α)(x-β)dx=-\frac{a}6(β-α)^3$$
for example
$$\frac16(-1-(-3)^2)=\frac43$$
