# Thread: integrating with respect to y / regions between curves

1. ## integrating with respect to y / regions between curves

Find the region R bounded by the graphs $\displaystyle y = x^3$ and$\displaystyle y = x+ 6$

The left curve $\displaystyle y=x+6$ becomes $\displaystyle x= y-6$
The right curve $\displaystyle y=x^3$ becomes$\displaystyle x=y^{1/3}$

How do I find the intersection point of the curve and the root which will be the pper and lower boundaries. The book says I can use synthetic division to find the roots.

2. ## Re: integrating with respect to y / regions between curves

The two curves will intersect when x^3 = x + 6. Thus, you must solve x^3 - x - 6 = 0. It is relatively easy to tell that 2 is a root of p(x) = x^3 - x - 6. Thus, by synthetic or long division we can factor p(x) as (x - 2)(x^2 + 2x + 3). Since x^2 + 2x + 3 does not factor, we have that p(x) has only one real root, which means that y = x^3 intersects y = x + 6 at exactly one point, which means that there is no region R bounded by the graphs.