y=x^3.
P = (3,27)
PQ is tangent to the curve at P
Picture of graph: http://i42.tinypic.com/fvl8gy.jpg
Find the area of the region enclosed between the curve, PQ and the x-axis.
I(x) = (1/4)x^4
I can see that x=0 at the origin...
I(0) = (1/4)(0)^4 = 0
How can I find point Q? It's tangent to the curve at point P...but I'm not sure how to find it :[
The area is 6.75..so how do I arrive at that? How can I find point Q? Thanks!
y=x^3 implies y'=3x^2 so at (3,27) y'=27
hence equation of PQ is y-27=27(x-3) and when y=0 we will have x=2 so Q is the point (2,0)
Area under curve is thus (1/4)x^4 evaluated from 0 to 3 =81/4 and the area under the line PQ is 27/2 so area between curve and line is 27/4=6.75