first find the tangent line
dy/dx=2x
at x=6, the slope is 2(6)=12
so
use point slope form
y-y1=m(x-x1)
point is (6,36)
so
y-36=12(x-6)
y-36=12x-72
y=12x-36
alright, so we know they intersect at x=6
and y=12x-36 is below y=x^2
so we do [tex] \int\limits^6_0 {x^2} \, dx - \int\limits^6_0 {12x-36} \, dx = \int\limits^6_0 {x^2-(12x-36)} \, dx = \int\limits^6_0 {x^2-12x+36} \, dx =[/tex]
[tex][\frac{x^3}{3}-6x^2+36x]\limits^6_0=(\frac{6^3}{3}-6(6)^2+36(6))-(0)=\frac{216}{3}-216+216=[/tex] [tex]72+0=72[/tex]
the area under the curve bounded by the lines and the x axis is 72 square units
