Hello, zer0e!
It should be simple Algebra . . .(okay, maybe not so simple).
Find the area of the plane region bounded by the curve $\displaystyle y\:=\:x^3$
and the tangent line to the curve at the point (1,1). Your tangent line is correct: .$\displaystyle y \:=\:3x  2$
To find the intersections, set the functions equal to each other:
. . $\displaystyle x^3 \:=\:3x  2 \quad\Rightarrow\quad x^3  3x + 2 \:=\:0$
We see that $\displaystyle x = 1$ is a solution, so $\displaystyle (x1)$ is a factor.
With long division: .$\displaystyle x^3  2x + 2 \;=\;(x1)(x^2+x2) \;=\;(x1)(x1)(x2)$
The intersections occurs at: .$\displaystyle (2,8),\;(1,1)$ Code:

 ...♥(1,1)
 ...o::*
     ..o:::::*    
o:::::::*
.:::::::*
o:::::::*
.::::::* 
o::::* 
:::* 
:* 
♥(2,8) 

Therefore: .$\displaystyle A \;=\;\int^1_{2}\bigg[x^3  (3x2)\bigg]\,dx$