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Math Help - Question on integration..finding the area?

  1. #1
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    Smile 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!
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  2. #2
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    Quote Originally Posted by rubbersoul923 View Post
    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 y = x^3 at the point P.

    so, find the equation of this tangent line, set y = 0 , and solve for x.
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  3. #3
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    Quote Originally Posted by rubbersoul923 View Post
    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
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  4. #4
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    You could also do this as a single integral by integrating with respect to y.

    When y= x^3, x= y^{1/3} and when [tex]y- 27= 27(x- 3), [tex]x= \frac{y+ 54}{27}, so the area is given by
    \int_{y= 0}^{27} \frac{y+ 54}{27}- y^{1/3} dy
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