1/2 = 7/14
3/7 = 6/14
7/14 + 6/14 = 13/14
Shaded Area: 1/14
Hope this helps! Message me with any questions!
6, -4, 2
You can get this by factoring the trinomial and setting each part, along with x-6, equal to 0. The trinomial can be factored to (x+4)(x-2)
Integrate by parts.
![\displaystyle \int 6x \sec(x) \tan(x) \, dx = uv - \int v\,du](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%206x%20%5Csec%28x%29%20%5Ctan%28x%29%20%5C%2C%20dx%20%3D%20uv%20-%20%5Cint%20v%5C%2Cdu)
with
![u = 6x \implies du = 6\,dx](https://tex.z-dn.net/?f=u%20%3D%206x%20%5Cimplies%20du%20%3D%206%5C%2Cdx)
![dv = \sec(x)\tan(x)\,dx \implies v = \sec(x)](https://tex.z-dn.net/?f=dv%20%3D%20%5Csec%28x%29%5Ctan%28x%29%5C%2Cdx%20%5Cimplies%20v%20%3D%20%5Csec%28x%29)
so that
![\displaystyle \int 6x \sec(x) \tan(x) \, dx = 6x\sec(x) - 6 \int \sec(x) \, dx \\\\ ~~~~~~~~ = \boxed{6x\sec(x) - 6 \ln|\sec(x) + \tan(x)| + C}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%206x%20%5Csec%28x%29%20%5Ctan%28x%29%20%5C%2C%20dx%20%3D%206x%5Csec%28x%29%20-%206%20%5Cint%20%5Csec%28x%29%20%5C%2C%20dx%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%20%5Cboxed%7B6x%5Csec%28x%29%20-%206%20%5Cln%7C%5Csec%28x%29%20%2B%20%5Ctan%28x%29%7C%20%2B%20C%7D)
where the last integral follows from
![\displaystyle \int \sec(x) \, dx = \int \frac{\sec(x)(\sec(x)+\tan(x))}{\sec(x)+\tan(x)} \, dx = \int \frac{d(\sec(x)+\tan(x))}{\sec(x)+\tan(x)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Csec%28x%29%20%5C%2C%20dx%20%3D%20%5Cint%20%5Cfrac%7B%5Csec%28x%29%28%5Csec%28x%29%2B%5Ctan%28x%29%29%7D%7B%5Csec%28x%29%2B%5Ctan%28x%29%7D%20%5C%2C%20dx%20%3D%20%5Cint%20%5Cfrac%7Bd%28%5Csec%28x%29%2B%5Ctan%28x%29%29%7D%7B%5Csec%28x%29%2B%5Ctan%28x%29%7D)