Definite Integration: Solving for k in 2k+4

[sallyj92]

The equation of the tangent at P is y=3 x - 2 . Let T be the region enclosed by
the graph of f , the tangent [PR] and the line x= k , between x = −2 and x= k
where − 2< k < 1.

Given that are of T is 2k+4, show that k satisfies the equation k^4-6k^2+8=0.

So I know you have to integrate x^3-3x-2 for the x=k and x=-2 but I struggled to integrate because I got k^4/4-3k^2/2-4k-10=0

I tried equating the integrated equation to 2k+4 but I didn't get the right answer, k^4-6k^2+8=0.

 

[Mark44]

Is there a drawing that goes with this problem? That would make it easier to understand what region they're talking about.

 

[sallyj92]

Is there a drawing that goes with this problem? That would make it easier to understand what region they're talking about.

[PLAIN]http://img97.imageshack.us/img97/8981/capturefal.jpg

 

[sallyj92]

[PLAIN]http://img97.imageshack.us/img97/8981/capturefal.jpg[/QUOTE]

https://mr-t-tes.wikispaces.com/file/view/Nov+2010+Mathematics+SL+paper+1+QP.pdf
Q10 - the whole question

 

[hunt_mat]

Now we can help you...

The area is given as the difference between the tangent and the function itself. So you integrate g(x)=f(x)-(3x-2)=x^3-3x+2 and you integrate this between -2 and k. This will give your relation if you set the area to be 2k+4