- #1
sallyj92
- 5
- 0
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.
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.