# Thread: Area bounded by curve and tangent line

1. ## Area bounded by curve and tangent line

Find the area of the plane region bounded by the curve y=x^3 and the tangent line to the curve at the point (1,1). So far, I have found the tangent line to be y = 3x - 2, and since the point x=1 was an intersection, it's one of the limits in the integral. However, I am having trouble finding the second intersection in order to know both the upper and lower limit. How would I calculate it?

2. Hello, zer0e!

It should be simple Algebra . . .(okay, maybe not so simple).

Find the area of the plane region bounded by the curve $y\:=\:x^3$
and the tangent line to the curve at the point (1,1).
Your tangent line is correct: . $y \:=\:3x - 2$

To find the intersections, set the functions equal to each other:
. . $x^3 \:=\:3x - 2 \quad\Rightarrow\quad x^3 - 3x + 2 \:=\:0$

We see that $x = 1$ is a solution, so $(x-1)$ is a factor.

With long division: . $x^3 - 2x + 2 \;=\;(x-1)(x^2+x-2) \;=\;(x-1)(x-1)(x-2)$

The intersections occurs at: . $(-2,-8),\;(1,1)$
Code:
              |
|      ...♥(1,1)
| ...o::*
- - - - - ..o:::::* - - - -
o::::|:::*
.::::::|:*
o:::::::*
.::::::* |
o::::*   |
:::*     |
:*       |
♥(-2,-8)  |
|

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

3. Ahhh I see. I had forgotten about the long division. Thanks!