Thread: Question on integration..finding the area?

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!

Originally Posted by rubbersoul923

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!

point Q is the x-intercept of the line tangent to at the point P.

so, find the equation of this tangent line, set y = 0 , and solve for x.

Originally Posted by rubbersoul923

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

You could also do this as a single integral by integrating with respect to y.

When , and when [tex]y- 27= 27(x- 3), [tex]x= \frac{y+ 54}{27}, so the area is given by