Area Between Y=x^3 & Its Tangent at x=1

[JOhnJDC]

Homework Statement

What is the area between y=x³ and its tangent at x=1

The Attempt at a Solution

The first derivative of y=x³, which is 3x², tells me that the slope of the tangent at x=1 is 3. That (1,1) is a point on the tangent line tells me that the equation of the tangent line is y=3x-2. Now, I know that I need to solve y=3x-2 and y=x³ simultaneously to find the points of intersection, and that the x coordinates of these points will be my limits of integration. However, I'm having trouble factoring x³-3x+2=0. Anyone have any tips/tricks for factoring an equation like this? I suppose my question has more to do with algebra than calculus.

Thanks.

 

[EvilKermit]

There's an easy way to solve the equation without having to do algebra. Assuming you have a graphing calculator you can just graph the two out and solve for the intersections.

However if you do not or need to use algebra, you can also use Synthetic division (a variation of long division of polynominals) if you don't know what it is this site exmplaisn it:
http://www.purplemath.com/modules/synthdiv.htm

This gives you (x-1)(x^2 + x -2 )
(x-1)(x-1)(x+2)
x = 1, -1, -2

 

[Paris.91]

The limits are 0 to 1... and the answer will be 0.5 sqr.units...

 

[JOhnJDC]

Thanks, EvilKermit. I used a graphing calculator to find the intersection at (-2,-8). I just wanted to do it algebraically. I'll check out the link you provided.

The limits are 0 to 1... and the answer will be 0.5 sqr.units...

y=x³ and y=3x-2 do not intersect at any point when x=0; nor is the area between the curves symmetrical about the y-axis. The limits of integration are -2 to 1, and integrating yields 27/4.

 

[EvilKermit]

Yea, I just realized my mistake above that x = 1, -2. There is no -1. The work is right above, just the answer was wrong at the end :)