Find the area between curve and straight line My first step is to let y of the first equation equal the second:

$2x -1 = x^2 + 6x + 2$

= $x^2 + 4x +3 = 0$

= $(x + 3)(x + 1)$

To find the area, I find the integral of the top most equation and subtract it from the integral of the bottom query, I believe that the top equation is the line so:

$\int_{-3}^{-1}2x-1 dx$

=> $x^2 - x$

=> $(1 -- 1) - (9 -- 3) = -12$

I now find the integral of the curve:

$\int_{-3}^{-1}x^2 + 6x + 2 dx$

=> $\frac{x^3}3 + 3x^2 + 2x$

=> ${(\frac{-1}{3} + 3 - 2}) - (-9 + 27 -6)$

=> $12\frac{1}{3}$

Then I subtract the first integral from the second:

$-12 - 12\frac{1}{3}$

This leads me to a minus number.  Where have I gone wrong?
Firstly,
$$[x^2-x]_{-3}^{-1}=(1-(-1))-(9-(-3))=2-12=-10$$

And,
$$(-\frac13-2)-(-9+27-6)=\frac23-12=-\frac{34}{3}$$

So, the result is
$$-10+\frac{34}{3}=\frac43$$
Firstly,
$$[x^2-x]_{-3}^{-1}=(1-(-1))-(9-(-3))=2-12=-10$$

And,
$$(-\frac13-2)-(-9+27-6)=\frac23-12=-\frac{34}{3}$$

So, the result is
$$-10+\frac{34}{3}=\frac43$$